{"id":317,"date":"2023-09-20T22:49:03","date_gmt":"2023-09-20T22:49:03","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/solving-optimization-problems\/"},"modified":"2024-08-05T13:30:33","modified_gmt":"2024-08-05T13:30:33","slug":"applied-optimization-problems-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/applied-optimization-problems-learn-it-1\/","title":{"raw":"Applied Optimization Problems: Learn It 1","rendered":"Applied Optimization Problems: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Tackle business problems to find the best ways to increase profits, minimize costs, or maximize revenue<\/li>\r\n\t<li>Use optimization methods to solve problems involving geometry<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solving Optimization Problems<\/h2>\r\n<p>A key application of calculus involves calculating a function's minimum or maximum value to optimize certain outcomes, such as reducing costs or increasing revenue. This is crucial in areas like manufacturing, where minimizing material usage for a specific product volume can be vital.<\/p>\r\n<h3>Solving Optimization Problems over a Closed, Bounded Interval<\/h3>\r\n<p id=\"fs-id1165043329454\"><strong>Optimization<\/strong> often involves finding the maximum or minimum value of a function under certain constraints. For example, consider maximizing the area of a rectangular garden while adhering to a fixed amount of available fencing. This requires adjusting the garden's dimensions within the set perimeter limits to find the optimal area.<\/p>\r\n<p>Here\u2019s an example of how we might approach this.<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165042623548\">A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides (Figure 1). Given [latex]100[\/latex] ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"462\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211219\/CNX_Calc_Figure_04_07_001.jpg\" alt=\"A drawing of a garden has x and y written on the vertical and horizontal sides, respectively. There is a rock wall running along the entire bottom horizontal length of the drawing.\" width=\"462\" height=\"378\" \/> Figure 1. We want to determine the measurements [latex]x[\/latex] and [latex]y[\/latex] that will create a garden with a maximum area using 100 ft of fencing.[\/caption]\r\n\r\n<p id=\"fs-id1165043262577\">Let [latex]x[\/latex] denote the length of the side of the garden perpendicular to the rock wall and [latex]y[\/latex] denote the length of the side parallel to the rock wall. Then the area of the garden is:<\/p>\r\n<div id=\"fs-id1165042515463\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=x\u00b7y[\/latex]<\/div>\r\n<p id=\"fs-id1165043222797\">We want to find the maximum possible area subject to the constraint that the total fencing is [latex]100[\/latex] ft. From Figure 1, the total amount of fencing used will be [latex]2x+y[\/latex].<\/p>\r\n<p>Therefore, the constraint equation is:<\/p>\r\n<div id=\"fs-id1165043249986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]2x+y=100[\/latex]<\/div>\r\n<p id=\"fs-id1165043272012\">Solving this equation for [latex]y[\/latex], we have [latex]y=100-2x[\/latex].<\/p>\r\n<p>Thus, we can write the area as:<\/p>\r\n<div id=\"fs-id1165043281405\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)=x \\cdot (100-2x)=100x-2x^2[\/latex]<\/div>\r\n<p id=\"fs-id1165043195368\">Before trying to maximize the area function [latex]A(x)=100x-2x^2[\/latex], we need to determine the domain under consideration.<\/p>\r\n<p>To construct a rectangular garden, we certainly need the lengths of both sides to be positive. Therefore, we need [latex]x&gt;0[\/latex] and [latex]y&gt;0[\/latex]. Since [latex]y=100-2x[\/latex], if [latex]y&gt;0[\/latex], then [latex]x&lt;50[\/latex].<\/p>\r\n<p>Therefore, we are trying to determine the maximum value of [latex]A(x)[\/latex] for [latex]x[\/latex] over the open interval [latex](0,50)[\/latex]. We do not know that a function necessarily has a maximum value over an open interval. However, we do know that a continuous function has an absolute maximum (and absolute minimum) over a closed interval.<\/p>\r\n<p>Therefore, let\u2019s consider the function [latex]A(x)=100x-2x^2[\/latex] over the closed interval [latex][0,50][\/latex]. If the maximum value occurs at an interior point, then we have found the value [latex]x[\/latex] in the open interval [latex](0,50)[\/latex] that maximizes the area of the garden.<\/p>\r\n<p>Therefore, we consider the following problem:<\/p>\r\n<p id=\"fs-id1165043094646\" style=\"text-align: center;\">Maximize [latex]A(x)=100x-2x^2[\/latex] over the interval [latex][0,50][\/latex]<\/p>\r\n<p id=\"fs-id1165042989293\">As mentioned earlier, since [latex]A[\/latex] is a continuous function on a closed, bounded interval, by the extreme value theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. At the endpoints, [latex]A(x)=0[\/latex]. Since the area is positive for all [latex]x[\/latex] in the open interval [latex](0,50)[\/latex], the maximum must occur at a critical point. Differentiating the function [latex]A(x)[\/latex], we obtain,<\/p>\r\n<div id=\"fs-id1165043380052\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A^{\\prime}(x)=100-4x[\/latex]<\/div>\r\n<p id=\"fs-id1165043178466\">Therefore, the only critical point is [latex]x=25[\/latex] (Figure 2). We conclude that the maximum area must occur when [latex]x=25[\/latex]. Then we have [latex]y=100-2x=100-2(25)=50[\/latex].<\/p>\r\n<p>To maximize the area of the garden, let [latex]x=25[\/latex] ft and [latex]y=50[\/latex] ft. The area of this garden is [latex]1250 \\, \\text{ft}^2[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"534\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211223\/CNX_Calc_Figure_04_07_002.jpg\" alt=\"The function A(x) = 100x \u2013 2x is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum area is 1250 square feet when x = 25 feet.\u201d\" width=\"534\" height=\"347\" \/> Figure 2. To maximize the area of the garden, we need to find the maximum value of the function [latex]A(x)=100x-2x^2[\/latex].[\/caption]\r\n<\/section>\r\n<section class=\"textbox watchIt\">\r\n<p>Watch the following video to see the worked solution to the example above.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=15&amp;end=180&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems15to180_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.7 Applied Optimization Problems\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165043035196\">Determine the maximum area if we want to make the same rectangular garden as in the example above, but we have [latex]200[\/latex] ft of fencing.<\/p>\r\n<p>[reveal-answer q=\"355002\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"355002\"]<\/p>\r\n<p>We need to maximize the function [latex]A(x)=200x-2x^2[\/latex] over the interval [latex][0,100][\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1165043102115\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043102115\"]<\/p>\r\n<p id=\"fs-id1165043102115\">The maximum area is [latex]5000 \\, \\text{ft}^2[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p id=\"fs-id1165042376423\">Here is a general strategy for solving optimization problems similar to these above.<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How To: Solve Optimization Problems<\/strong><\/p>\r\n<ol id=\"fs-id1165043013367\">\r\n\t<li><strong>Identify All Variables:<\/strong> If applicable, sketch the problem scenario and label all variables.<\/li>\r\n\t<li><strong>Determine Objective:<\/strong> Identify which quantity needs to be maximized or minimized, and specify the range of values for any other relevant variables.<\/li>\r\n\t<li><strong>Develop a Formula:<\/strong> Write a formula for the objective quantity in terms of the variables, which may involve multiple variables.<\/li>\r\n\t<li><strong>Formulate Equations:<\/strong> Relate independent variables with any equations necessary to express the objective quantity as a function of one variable.<\/li>\r\n\t<li><strong>Set Domain:<\/strong> Determine the domain of consideration for the function based on the practical constraints of the problem.<\/li>\r\n\t<li><strong>Find Extremes:<\/strong> Calculate the maximum or minimum value of the function, typically by identifying critical points and evaluating the function at endpoints.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<p id=\"fs-id1165043029463\">Let\u2019s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be used.<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165042638735\">An open-top box is to be made from a [latex]24[\/latex] in. by [latex]36[\/latex] in. piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume?<\/p>\r\n<p>[reveal-answer q=\"fs-id1165043092635\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043092635\"]<\/p>\r\n<p id=\"fs-id1165043092635\"><strong>Step 1<\/strong>: Let [latex]x[\/latex] be the side length of the square to be removed from each corner (Figure 3). Then, the remaining four flaps can be folded up to form an open-top box. Let [latex]V[\/latex] be the volume of the resulting box.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"599\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211226\/CNX_Calc_Figure_04_07_003.jpg\" alt=\"There are two figures for this figure. The first one is a rectangle with sides 24 in and 36 in, with each corner having a square of side length x taken out of it. In the second picture, there is a box with side lengths x in, 24 \u2013 2x in, and 36 \u2013 2x in.\" width=\"599\" height=\"333\" \/> Figure 3. A square with side length [latex]x[\/latex] inches is removed from each corner of the piece of cardboard. The remaining flaps are folded to form an open-top box.[\/caption]\r\n\r\n<p id=\"fs-id1165042330556\"><strong>Step 2<\/strong>: We are trying to maximize the volume of a box. Therefore, the problem is to maximize [latex]V[\/latex].<\/p>\r\n<p id=\"fs-id1165043251052\"><strong>Step 3<\/strong>: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is [latex]V=L \\cdot W \\cdot H[\/latex], where [latex]L, \\, W[\/latex], and [latex]H[\/latex] are the length, width, and height, respectively.<\/p>\r\n<p><strong>Step 4<\/strong>: From Figure 3, we see that the height of the box is [latex]x[\/latex] inches, the length is [latex]36-2x[\/latex] inches, and the width is [latex]24-2x[\/latex] inches. Therefore, the volume of the box is,<\/p>\r\n<div id=\"fs-id1165042645725\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(x)=(36-2x)(24-2x)x=4x^3-120x^2+864x[\/latex]<\/div>\r\n<p id=\"fs-id1165043115396\"><strong>Step 5<\/strong>: To determine the domain of consideration, let\u2019s examine Figure 3. Certainly, we need [latex]x&gt;0[\/latex]. Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, 24 in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for [latex]x[\/latex] over the open interval [latex](0,12)[\/latex]. Since [latex]V[\/latex] is a continuous function over the closed interval [latex][0,12][\/latex], we know [latex]V[\/latex] will have an absolute maximum over the closed interval. Therefore, we consider [latex]V[\/latex] over the closed interval [latex][0,12][\/latex] and check whether the absolute maximum occurs at an interior point.<\/p>\r\n<p id=\"fs-id1165043017956\"><strong>Step 6<\/strong>: Since [latex]V(x)[\/latex] is a continuous function over the closed, bounded interval [latex][0,12][\/latex], [latex]V[\/latex] must have an absolute maximum (and an absolute minimum). Since [latex]V(x)=0[\/latex] at the endpoints and [latex]V(x)&gt;0[\/latex] for [latex]0&lt;x&lt;12[\/latex], the maximum must occur at a critical point. The derivative is:<\/p>\r\n<div id=\"fs-id1165043333862\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V^{\\prime}(x)=12x^2-240x+864[\/latex]<\/div>\r\n<p id=\"fs-id1165042608183\">To find the critical points, we need to solve the equation:<\/p>\r\n<div id=\"fs-id1165042946835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]12x^2-240x+864=0[\/latex]<\/div>\r\n<p id=\"fs-id1165043209010\">Dividing both sides of this equation by [latex]12[\/latex], the problem simplifies to solving the equation:<\/p>\r\n<div id=\"fs-id1165042982075\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x^2-20x+72=0[\/latex]<\/div>\r\n<p id=\"fs-id1165043060298\">Using the quadratic formula, we find that the critical points are,<\/p>\r\n<div id=\"fs-id1165042367652\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=\\frac{20 \\pm \\sqrt{(-20)^2-4(1)(72)}}{2}=\\frac{20 \\pm \\sqrt{112}}{2}=\\frac{20 \\pm 4\\sqrt{7}}{2}=10 \\pm 2\\sqrt{7}[\/latex]<\/div>\r\n<p id=\"fs-id1165043263257\">Since [latex]10+2\\sqrt{7}[\/latex] is not in the domain of consideration, the only critical point we need to consider is [latex]10-2\\sqrt{7}[\/latex]. Therefore, the volume is maximized if we let [latex]x=10-2\\sqrt{7}[\/latex] in.<\/p>\r\n<p>The maximum volume is [latex]V(10-2\\sqrt{7})=640+448\\sqrt{7}\\approx 1825 \\, \\text{in}^3[\/latex] as shown in the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"719\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211230\/CNX_Calc_Figure_04_07_004.jpg\" alt=\"The function V(x) = 4x3 \u2013 120x2 + 864x is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum volume is approximately 1825 cubic inches when x \u2248 4.7 inches.\u201d\" width=\"719\" height=\"388\" \/> Figure 4. Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial.[\/caption]\r\n\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox interact\">\r\n<p><a href=\"https:\/\/www.khanacademy.org\/math\/old-differential-calculus\/derivative-applications-dc\/optimization-dc\/v\/optimizing-box-volume-graphically\" target=\"_blank\" rel=\"noopener\">Watch a video about optimizing the volume of a box.<\/a><\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165043329514\">An island is [latex]2[\/latex] mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is [latex]6[\/latex] mi west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of [latex]8[\/latex] mph and swims at a rate of [latex]3[\/latex] mph. How far should the visitor run before swimming to minimize the time it takes to reach the island?<\/p>\r\n<p>[reveal-answer q=\"fs-id1165043276181\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043276181\"]<\/p>\r\n<p id=\"fs-id1165043276181\"><strong>Step 1:<\/strong> Let [latex]x[\/latex] be the distance running and let [latex]y[\/latex] be the distance swimming (Figure 5). Let [latex]T[\/latex] be the time it takes to get from the cabin to the island.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"594\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211233\/CNX_Calc_Figure_04_07_011.jpg\" alt=\"The cabin is x miles from the shore. From that point on the shore, the island is y miles away. If you were to continue the line from the cabin to the shore (the x miles one) and if you were to draw a line from the island parallel to the shore, then the lines would extend 2 miles from the island and 6 miles from the cabin before intersecting.\" width=\"594\" height=\"352\" \/> Figure 5. How can we choose [latex]x[\/latex] and [latex]y[\/latex] to minimize the travel time from the cabin to the island?[\/caption]\r\n\r\n<p id=\"fs-id1165042330905\"><strong>Step 2<\/strong>: The problem is to minimize [latex]T[\/latex].<\/p>\r\n<p id=\"fs-id1165043346985\"><strong>Step 3<\/strong>: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent swimming. Since Distance [latex]=[\/latex] Rate [latex]\\times[\/latex] Time [latex](D=R \\times T)[\/latex], the time spent running is<\/p>\r\n<div id=\"fs-id1165043431452\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T_{\\text{running}}=\\frac{D_{\\text{running}}}{R_{\\text{running}}}=\\frac{x}{8}[\/latex],<\/div>\r\n<p>and the time spent swimming is,<\/p>\r\n<div id=\"fs-id1165042369650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T_{\\text{swimming}}=\\frac{D_{\\text{swimming}}}{R_{\\text{swimming}}}=\\frac{y}{3}[\/latex]<\/div>\r\n<p id=\"fs-id1165043262597\">Therefore, the total time spent traveling is,<\/p>\r\n<div id=\"fs-id1165043433402\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=\\frac{x}{8}+\\frac{y}{3}[\/latex]<\/div>\r\n<p id=\"fs-id1165042517820\"><strong>Step 4<\/strong>: From Figure 5, the line segment of [latex]y[\/latex] miles forms the hypotenuse of a right triangle with legs of length [latex]2[\/latex] mi and [latex]6-x[\/latex] mi. Therefore, by the Pythagorean theorem, [latex]2^2+(6-x)^2=y^2[\/latex], and we obtain [latex]y=\\sqrt{(6-x)^2+4}[\/latex]. Thus, the total time spent traveling is given by the function<\/p>\r\n<div id=\"fs-id1165043430018\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T(x)=\\frac{x}{8}+\\frac{\\sqrt{(6-x)^2+4}}{3}[\/latex]<\/div>\r\n<p id=\"fs-id1165042328992\"><strong>Step 5<\/strong>: From Figure 5, we see that [latex]0\\le x\\le 6[\/latex]. Therefore, [latex][0,6][\/latex] is the domain of consideration.<\/p>\r\n<p id=\"fs-id1165043066617\"><strong>Step 6<\/strong>: Since [latex]T(x)[\/latex] is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let\u2019s begin by looking for any critical points of [latex]T[\/latex] over the interval [latex][0,6][\/latex]. The derivative is,<\/p>\r\n<div id=\"fs-id1165043353795\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T^{\\prime}(x)=\\frac{1}{8}-\\frac{1}{2}\\frac{[(6-x)^2+4]^{-1\/2}}{3} \\cdot 2(6-x)=\\frac{1}{8}-\\frac{(6-x)}{3\\sqrt{(6-x)^2+4}}[\/latex]<\/div>\r\n<p id=\"fs-id1165042516379\">If [latex]T^{\\prime}(x)=0[\/latex], then,<\/p>\r\n<div id=\"fs-id1165042705986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{8}=\\frac{6-x}{3\\sqrt{(6-x)^2+4}}[\/latex]<\/div>\r\n<p id=\"fs-id1165043107403\">Therefore,<\/p>\r\n<div id=\"fs-id1165042343282\" class=\"equation\" style=\"text-align: center;\">[latex]3\\sqrt{(6-x)^2+4}=8(6-x)[\/latex]<\/div>\r\n<p id=\"fs-id1165043284394\">Squaring both sides of this equation, we see that if [latex]x[\/latex] satisfies this equation, then [latex]x[\/latex] must satisfy,<\/p>\r\n<div id=\"fs-id1165043103774\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]9[(6-x)^2+4]=64(6-x)^2[\/latex],<\/div>\r\n<p id=\"fs-id1165042923339\">which implies,<\/p>\r\n<div id=\"fs-id1165043285288\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]55(6-x)^2=36[\/latex]<\/div>\r\n<p id=\"fs-id1165042711030\">We conclude that if [latex]x[\/latex] is a critical point, then [latex]x[\/latex] satisfies,<\/p>\r\n<div id=\"fs-id1165042923510\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](x-6)^2=\\frac{36}{55}[\/latex]<\/div>\r\n<p id=\"fs-id1165043396085\">Therefore, the possibilities for critical points are,<\/p>\r\n<div id=\"fs-id1165042350053\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=6 \\pm \\frac{6}{\\sqrt{55}}[\/latex]<\/div>\r\n<p id=\"fs-id1165042616899\">Since [latex]x=6+\\frac{6}{\\sqrt{55}}[\/latex] is not in the domain, it is not a possibility for a critical point. On the other hand, [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] is in the domain.<\/p>\r\n<p>Since we squared both sides of the equation to arrive at the possible critical points, it remains to verify that it is satisfied by [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex]. Since [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] does satisfy that equation, we conclude that [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] is a critical point, and it is the only one.<\/p>\r\n<p>To justify that the time is minimized for this value of [latex]x[\/latex], we just need to check the values of [latex]T(x)[\/latex] at the endpoints [latex]x=0[\/latex] and [latex]x=6[\/latex], and compare them with the value of [latex]T(x)[\/latex] at the critical point [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex]. We find that [latex]T(0)\\approx 2.108[\/latex] h and [latex]T(6)\\approx 1.417[\/latex] h, whereas [latex]T(6-\\frac{6}{\\sqrt{55}})\\approx 1.368[\/latex] h.<\/p>\r\n<p>Therefore, we conclude that [latex]T[\/latex] has a local minimum at [latex]x\\approx 5.19[\/latex] mi.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>In business, companies are interested in <span class=\"no-emphasis\">maximizing revenue<\/span>. In the following example, we consider a scenario in which a company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a car. Let\u2019s use these data to determine the price the company should charge to maximize the amount of money it brings in.<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165042515832\">Owners of a car rental company have determined that if they charge customers [latex]p[\/latex] dollars per day to rent a car, where [latex]50\\le p\\le 200[\/latex], the number of cars [latex]n[\/latex] they rent per day can be modeled by the linear function [latex]n(p)=1000-5p[\/latex]. If they charge [latex]$50[\/latex] per day or less, they will rent all their cars. If they charge [latex]$200[\/latex] per day or more, they will not rent any cars. Assuming the owners plan to charge customers between [latex]$50[\/latex] per day and [latex]$200[\/latex] per day to rent a car, how much should they charge to maximize their revenue?<\/p>\r\n<p>[reveal-answer q=\"fs-id1165043423441\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043423441\"]<\/p>\r\n<p id=\"fs-id1165043423441\"><strong>Step 1<\/strong>: Let [latex]p[\/latex] be the price charged per car per day and let [latex]n[\/latex] be the number of cars rented per day. Let [latex]R[\/latex] be the revenue per day.<\/p>\r\n<p id=\"fs-id1165043431290\"><strong>Step 2<\/strong>: The problem is to maximize [latex]R[\/latex].<\/p>\r\n<p id=\"fs-id1165043087644\"><strong>Step 3<\/strong>: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day\u2014that is, [latex]R=n \\times p[\/latex].<\/p>\r\n<p id=\"fs-id1165042514435\"><strong>Step 4<\/strong>: Since the number of cars rented per day is modeled by the linear function [latex]n(p)=1000-5p[\/latex], the revenue [latex]R[\/latex] can be represented by the function:<\/p>\r\n<div id=\"fs-id1165043256835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(p)=n \\times p=(1000-5p)p=-5p^2+1000p[\/latex]<\/div>\r\n<p id=\"fs-id1165042406610\"><strong>Step 5<\/strong>: Since the owners plan to charge between [latex]$50[\/latex] per car per day and [latex]$200[\/latex] per car per day, the problem is to find the maximum revenue [latex]R(p)[\/latex] for [latex]p[\/latex] in the closed interval [latex][50,200][\/latex].<\/p>\r\n<p id=\"fs-id1165043173962\"><strong>Step 6<\/strong>: Since [latex]R[\/latex] is a continuous function over the closed, bounded interval [latex][50,200][\/latex], it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is [latex]R^{\\prime}(p)=-10p+1000[\/latex]. Therefore, the critical point is [latex]p=100[\/latex].<\/p>\r\n<p>When [latex]p=100[\/latex], [latex]R(100)=$50,000[\/latex]. When [latex]p=50[\/latex], [latex]R(p)=$37,500[\/latex]. When [latex]p=200[\/latex], [latex]R(p)=$0[\/latex]. Therefore, the absolute maximum occurs at [latex]p=$100[\/latex].<\/p>\r\n<p>The car rental company should charge [latex]$100[\/latex] per day per car to maximize revenue as shown in the following figure.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"495\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211236\/CNX_Calc_Figure_04_07_010.jpg\" alt=\"The function R(p) is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum revenue is 50,000 per day when the price charged per car is 100 per day.\u201d\" width=\"495\" height=\"270\" \/> Figure 6. To maximize revenue, a car rental company has to balance the price of a rental against the number of cars people will rent at that price.[\/caption]\r\n\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=913&amp;end=1024&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems913to1024_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.7 Applied Optimization Problems\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165042456921\">A rectangle is to be inscribed in the ellipse<\/p>\r\n<div id=\"fs-id1165042456924\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{x^2}{4}+y^2=1[\/latex]<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1165042645676\">What should the dimensions of the rectangle be to maximize its area? What is the maximum area?<\/p>\r\n<p>[reveal-answer q=\"fs-id1165042519826\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042519826\"]<\/p>\r\n<p id=\"fs-id1165042519826\"><strong>Step 1<\/strong>: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let [latex]L[\/latex] be the length of the rectangle and [latex]W[\/latex] be its width. Let [latex]A[\/latex] be the area of the rectangle.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"575\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211239\/CNX_Calc_Figure_04_07_007.jpg\" alt=\"The ellipse x2\/4 + y2 = 1 is drawn with its x intercepts being \u00b12 and its y intercepts being \u00b11. There is a rectangle inscribed in the ellipse with length L (in the x-direction) and width W.\" width=\"575\" height=\"350\" \/> Figure 7. We want to maximize the area of a rectangle inscribed in an ellipse.[\/caption]\r\n\r\n<p id=\"fs-id1165042371013\"><strong>Step 2<\/strong>: The problem is to maximize [latex]A[\/latex].<\/p>\r\n<p id=\"fs-id1165042377481\"><strong>Step 3<\/strong>: The area of the rectangle is [latex]A=L \\times W[\/latex].<\/p>\r\n<p id=\"fs-id1165043257695\"><strong>Step 4<\/strong>: Let [latex](x,y)[\/latex] be the corner of the rectangle that lies in the first quadrant, as shown in Figure 7. We can write length [latex]L=2x[\/latex] and width [latex]W=2y[\/latex]. Since [latex]\\frac{x^2}{4}+y^2=1[\/latex] and [latex]y&gt;0[\/latex], we have [latex]y=\\sqrt{1 - \\frac{x^2}{4}}[\/latex]. Therefore, the area is:<\/p>\r\n<div id=\"fs-id1165042320390\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=L \\times W=(2x)(2y)=4x\\sqrt{1 - \\frac{x^2}{4}}=2x\\sqrt{4-x^2}[\/latex]<\/div>\r\n<p id=\"fs-id1165043380076\"><strong>Step 5<\/strong>: From Figure 7, we see that to inscribe a rectangle in the ellipse, the [latex]x[\/latex]-coordinate of the corner in the first quadrant must satisfy [latex]0&lt;x&lt;2[\/latex]. Therefore, the problem reduces to looking for the maximum value of [latex]A(x)[\/latex] over the open interval [latex](0,2)[\/latex].<\/p>\r\n<p>Since [latex]A(x)[\/latex] will have an absolute maximum (and absolute minimum) over the closed interval [latex][0,2][\/latex], we consider [latex]A(x)=2x\\sqrt{4-x^2}[\/latex] over the interval [latex][0,2][\/latex]. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval.<\/p>\r\n<p id=\"fs-id1165042331466\"><strong>Step 6<\/strong>: As mentioned earlier, [latex]A(x)[\/latex] is a continuous function over the closed, bounded interval [latex][0,2][\/latex]. Therefore, it has an absolute maximum (and absolute minimum). At the endpoints [latex]x=0[\/latex] and [latex]x=2[\/latex], [latex]A(x)=0[\/latex]. For [latex]0&lt;x&lt;2[\/latex], [latex]A(x)&gt;0[\/latex]. Therefore, the maximum must occur at a critical point. Taking the derivative of [latex]A(x)[\/latex], we obtain,<\/p>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} A^{\\prime}(x) &amp; =2\\sqrt{4-x^2}+2x \\cdot \\frac{1}{2\\sqrt{4-x^2}}(-2x) \\\\ &amp; =2\\sqrt{4-x^2}-\\frac{2x^2}{\\sqrt{4-x^2}} \\\\ &amp; =\\frac{8-4x^2}{\\sqrt{4-x^2}} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165043379942\">To find critical points, we need to find where [latex]A^{\\prime}(x)=0[\/latex]. We can see that if [latex]x[\/latex] is a solution of<\/p>\r\n<div id=\"fs-id1165043354675\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{8-4x^2}{\\sqrt{4-x^2}}=0[\/latex],<\/div>\r\n<p id=\"fs-id1165043183716\">then [latex]x[\/latex] must satisfy<\/p>\r\n<div id=\"fs-id1165043281800\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]8-4x^2=0[\/latex]<\/div>\r\n<p id=\"fs-id1165042515040\">Therefore, [latex]x^2=2[\/latex]. Thus, [latex]x=\\pm \\sqrt{2}[\/latex] are the possible solutions of[latex]A^{\\prime}(x)[\/latex].<\/p>\r\n<p>Since we are considering [latex]x[\/latex] over the interval [latex][0,2][\/latex], [latex]x=\\sqrt{2}[\/latex] is a possibility for a critical point, but [latex]x=\u2212\\sqrt{2}[\/latex] is not. Therefore, we check whether [latex]\\sqrt{2}[\/latex] is a solution of [latex]A^{\\prime}(x)[\/latex]. Since [latex]x=\\sqrt{2}[\/latex] is a solution of[latex]A^{\\prime}(x)[\/latex], we conclude that [latex]\\sqrt{2}[\/latex] is the only critical point of [latex]A(x)[\/latex] in the interval [latex][0,2][\/latex].<\/p>\r\n<p>Therefore, [latex]A(x)[\/latex] must have an absolute maximum at the critical point [latex]x=\\sqrt{2}[\/latex]. To determine the dimensions of the rectangle, we need to find the length [latex]L[\/latex] and the width [latex]W[\/latex]. If [latex]x=\\sqrt{2}[\/latex] then,<\/p>\r\n<div id=\"fs-id1165042708515\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\sqrt{1-\\frac{(\\sqrt{2})^2}{4}}=\\sqrt{1-\\frac{1}{2}}=\\frac{1}{\\sqrt{2}}[\/latex]<\/div>\r\n<p id=\"fs-id1165043376531\">Therefore, the dimensions of the rectangle are [latex]L=2x=2\\sqrt{2}[\/latex] and [latex]W=2y=\\frac{2}{\\sqrt{2}}=\\sqrt{2}[\/latex]. The area of this rectangle is [latex]A=L \\times W=(2\\sqrt{2})(\\sqrt{2})=4[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288427[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Tackle business problems to find the best ways to increase profits, minimize costs, or maximize revenue<\/li>\n<li>Use optimization methods to solve problems involving geometry<\/li>\n<\/ul>\n<\/section>\n<h2>Solving Optimization Problems<\/h2>\n<p>A key application of calculus involves calculating a function&#8217;s minimum or maximum value to optimize certain outcomes, such as reducing costs or increasing revenue. This is crucial in areas like manufacturing, where minimizing material usage for a specific product volume can be vital.<\/p>\n<h3>Solving Optimization Problems over a Closed, Bounded Interval<\/h3>\n<p id=\"fs-id1165043329454\"><strong>Optimization<\/strong> often involves finding the maximum or minimum value of a function under certain constraints. For example, consider maximizing the area of a rectangular garden while adhering to a fixed amount of available fencing. This requires adjusting the garden&#8217;s dimensions within the set perimeter limits to find the optimal area.<\/p>\n<p>Here\u2019s an example of how we might approach this.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1165042623548\">A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides (Figure 1). Given [latex]100[\/latex] ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?<\/p>\n<figure style=\"width: 462px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211219\/CNX_Calc_Figure_04_07_001.jpg\" alt=\"A drawing of a garden has x and y written on the vertical and horizontal sides, respectively. There is a rock wall running along the entire bottom horizontal length of the drawing.\" width=\"462\" height=\"378\" \/><figcaption class=\"wp-caption-text\">Figure 1. We want to determine the measurements [latex]x[\/latex] and [latex]y[\/latex] that will create a garden with a maximum area using 100 ft of fencing.<\/figcaption><\/figure>\n<p id=\"fs-id1165043262577\">Let [latex]x[\/latex] denote the length of the side of the garden perpendicular to the rock wall and [latex]y[\/latex] denote the length of the side parallel to the rock wall. Then the area of the garden is:<\/p>\n<div id=\"fs-id1165042515463\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=x\u00b7y[\/latex]<\/div>\n<p id=\"fs-id1165043222797\">We want to find the maximum possible area subject to the constraint that the total fencing is [latex]100[\/latex] ft. From Figure 1, the total amount of fencing used will be [latex]2x+y[\/latex].<\/p>\n<p>Therefore, the constraint equation is:<\/p>\n<div id=\"fs-id1165043249986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]2x+y=100[\/latex]<\/div>\n<p id=\"fs-id1165043272012\">Solving this equation for [latex]y[\/latex], we have [latex]y=100-2x[\/latex].<\/p>\n<p>Thus, we can write the area as:<\/p>\n<div id=\"fs-id1165043281405\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A(x)=x \\cdot (100-2x)=100x-2x^2[\/latex]<\/div>\n<p id=\"fs-id1165043195368\">Before trying to maximize the area function [latex]A(x)=100x-2x^2[\/latex], we need to determine the domain under consideration.<\/p>\n<p>To construct a rectangular garden, we certainly need the lengths of both sides to be positive. Therefore, we need [latex]x>0[\/latex] and [latex]y>0[\/latex]. Since [latex]y=100-2x[\/latex], if [latex]y>0[\/latex], then [latex]x<50[\/latex].<\/p>\n<p>Therefore, we are trying to determine the maximum value of [latex]A(x)[\/latex] for [latex]x[\/latex] over the open interval [latex](0,50)[\/latex]. We do not know that a function necessarily has a maximum value over an open interval. However, we do know that a continuous function has an absolute maximum (and absolute minimum) over a closed interval.<\/p>\n<p>Therefore, let\u2019s consider the function [latex]A(x)=100x-2x^2[\/latex] over the closed interval [latex][0,50][\/latex]. If the maximum value occurs at an interior point, then we have found the value [latex]x[\/latex] in the open interval [latex](0,50)[\/latex] that maximizes the area of the garden.<\/p>\n<p>Therefore, we consider the following problem:<\/p>\n<p id=\"fs-id1165043094646\" style=\"text-align: center;\">Maximize [latex]A(x)=100x-2x^2[\/latex] over the interval [latex][0,50][\/latex]<\/p>\n<p id=\"fs-id1165042989293\">As mentioned earlier, since [latex]A[\/latex] is a continuous function on a closed, bounded interval, by the extreme value theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. At the endpoints, [latex]A(x)=0[\/latex]. Since the area is positive for all [latex]x[\/latex] in the open interval [latex](0,50)[\/latex], the maximum must occur at a critical point. Differentiating the function [latex]A(x)[\/latex], we obtain,<\/p>\n<div id=\"fs-id1165043380052\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A^{\\prime}(x)=100-4x[\/latex]<\/div>\n<p id=\"fs-id1165043178466\">Therefore, the only critical point is [latex]x=25[\/latex] (Figure 2). We conclude that the maximum area must occur when [latex]x=25[\/latex]. Then we have [latex]y=100-2x=100-2(25)=50[\/latex].<\/p>\n<p>To maximize the area of the garden, let [latex]x=25[\/latex] ft and [latex]y=50[\/latex] ft. The area of this garden is [latex]1250 \\, \\text{ft}^2[\/latex].<\/p>\n<figure style=\"width: 534px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211223\/CNX_Calc_Figure_04_07_002.jpg\" alt=\"The function A(x) = 100x \u2013 2x is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum area is 1250 square feet when x = 25 feet.\u201d\" width=\"534\" height=\"347\" \/><figcaption class=\"wp-caption-text\">Figure 2. To maximize the area of the garden, we need to find the maximum value of the function [latex]A(x)=100x-2x^2[\/latex].<\/figcaption><\/figure>\n<\/section>\n<section class=\"textbox watchIt\">\n<p>Watch the following video to see the worked solution to the example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=15&amp;end=180&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/button><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems15to180_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.7 Applied Optimization Problems&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1165043035196\">Determine the maximum area if we want to make the same rectangular garden as in the example above, but we have [latex]200[\/latex] ft of fencing.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q355002\">Hint<\/button><\/p>\n<div id=\"q355002\" class=\"hidden-answer\" style=\"display: none\">\n<p>We need to maximize the function [latex]A(x)=200x-2x^2[\/latex] over the interval [latex][0,100][\/latex].<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043102115\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043102115\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043102115\">The maximum area is [latex]5000 \\, \\text{ft}^2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1165042376423\">Here is a general strategy for solving optimization problems similar to these above.<\/p>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Solve Optimization Problems<\/strong><\/p>\n<ol id=\"fs-id1165043013367\">\n<li><strong>Identify All Variables:<\/strong> If applicable, sketch the problem scenario and label all variables.<\/li>\n<li><strong>Determine Objective:<\/strong> Identify which quantity needs to be maximized or minimized, and specify the range of values for any other relevant variables.<\/li>\n<li><strong>Develop a Formula:<\/strong> Write a formula for the objective quantity in terms of the variables, which may involve multiple variables.<\/li>\n<li><strong>Formulate Equations:<\/strong> Relate independent variables with any equations necessary to express the objective quantity as a function of one variable.<\/li>\n<li><strong>Set Domain:<\/strong> Determine the domain of consideration for the function based on the practical constraints of the problem.<\/li>\n<li><strong>Find Extremes:<\/strong> Calculate the maximum or minimum value of the function, typically by identifying critical points and evaluating the function at endpoints.<\/li>\n<\/ol>\n<\/section>\n<p id=\"fs-id1165043029463\">Let\u2019s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be used.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1165042638735\">An open-top box is to be made from a [latex]24[\/latex] in. by [latex]36[\/latex] in. piece of cardboard by removing a square from each corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get a box with the maximum volume?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043092635\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043092635\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043092635\"><strong>Step 1<\/strong>: Let [latex]x[\/latex] be the side length of the square to be removed from each corner (Figure 3). Then, the remaining four flaps can be folded up to form an open-top box. Let [latex]V[\/latex] be the volume of the resulting box.<\/p>\n<figure style=\"width: 599px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211226\/CNX_Calc_Figure_04_07_003.jpg\" alt=\"There are two figures for this figure. The first one is a rectangle with sides 24 in and 36 in, with each corner having a square of side length x taken out of it. In the second picture, there is a box with side lengths x in, 24 \u2013 2x in, and 36 \u2013 2x in.\" width=\"599\" height=\"333\" \/><figcaption class=\"wp-caption-text\">Figure 3. A square with side length [latex]x[\/latex] inches is removed from each corner of the piece of cardboard. The remaining flaps are folded to form an open-top box.<\/figcaption><\/figure>\n<p id=\"fs-id1165042330556\"><strong>Step 2<\/strong>: We are trying to maximize the volume of a box. Therefore, the problem is to maximize [latex]V[\/latex].<\/p>\n<p id=\"fs-id1165043251052\"><strong>Step 3<\/strong>: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is [latex]V=L \\cdot W \\cdot H[\/latex], where [latex]L, \\, W[\/latex], and [latex]H[\/latex] are the length, width, and height, respectively.<\/p>\n<p><strong>Step 4<\/strong>: From Figure 3, we see that the height of the box is [latex]x[\/latex] inches, the length is [latex]36-2x[\/latex] inches, and the width is [latex]24-2x[\/latex] inches. Therefore, the volume of the box is,<\/p>\n<div id=\"fs-id1165042645725\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(x)=(36-2x)(24-2x)x=4x^3-120x^2+864x[\/latex]<\/div>\n<p id=\"fs-id1165043115396\"><strong>Step 5<\/strong>: To determine the domain of consideration, let\u2019s examine Figure 3. Certainly, we need [latex]x>0[\/latex]. Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, 24 in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there is a maximum volume of the box for [latex]x[\/latex] over the open interval [latex](0,12)[\/latex]. Since [latex]V[\/latex] is a continuous function over the closed interval [latex][0,12][\/latex], we know [latex]V[\/latex] will have an absolute maximum over the closed interval. Therefore, we consider [latex]V[\/latex] over the closed interval [latex][0,12][\/latex] and check whether the absolute maximum occurs at an interior point.<\/p>\n<p id=\"fs-id1165043017956\"><strong>Step 6<\/strong>: Since [latex]V(x)[\/latex] is a continuous function over the closed, bounded interval [latex][0,12][\/latex], [latex]V[\/latex] must have an absolute maximum (and an absolute minimum). Since [latex]V(x)=0[\/latex] at the endpoints and [latex]V(x)>0[\/latex] for [latex]0<x<12[\/latex], the maximum must occur at a critical point. The derivative is:<\/p>\n<div id=\"fs-id1165043333862\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V^{\\prime}(x)=12x^2-240x+864[\/latex]<\/div>\n<p id=\"fs-id1165042608183\">To find the critical points, we need to solve the equation:<\/p>\n<div id=\"fs-id1165042946835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]12x^2-240x+864=0[\/latex]<\/div>\n<p id=\"fs-id1165043209010\">Dividing both sides of this equation by [latex]12[\/latex], the problem simplifies to solving the equation:<\/p>\n<div id=\"fs-id1165042982075\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x^2-20x+72=0[\/latex]<\/div>\n<p id=\"fs-id1165043060298\">Using the quadratic formula, we find that the critical points are,<\/p>\n<div id=\"fs-id1165042367652\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=\\frac{20 \\pm \\sqrt{(-20)^2-4(1)(72)}}{2}=\\frac{20 \\pm \\sqrt{112}}{2}=\\frac{20 \\pm 4\\sqrt{7}}{2}=10 \\pm 2\\sqrt{7}[\/latex]<\/div>\n<p id=\"fs-id1165043263257\">Since [latex]10+2\\sqrt{7}[\/latex] is not in the domain of consideration, the only critical point we need to consider is [latex]10-2\\sqrt{7}[\/latex]. Therefore, the volume is maximized if we let [latex]x=10-2\\sqrt{7}[\/latex] in.<\/p>\n<p>The maximum volume is [latex]V(10-2\\sqrt{7})=640+448\\sqrt{7}\\approx 1825 \\, \\text{in}^3[\/latex] as shown in the following graph.<\/p>\n<figure style=\"width: 719px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211230\/CNX_Calc_Figure_04_07_004.jpg\" alt=\"The function V(x) = 4x3 \u2013 120x2 + 864x is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum volume is approximately 1825 cubic inches when x \u2248 4.7 inches.\u201d\" width=\"719\" height=\"388\" \/><figcaption class=\"wp-caption-text\">Figure 4. Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox interact\">\n<p><a href=\"https:\/\/www.khanacademy.org\/math\/old-differential-calculus\/derivative-applications-dc\/optimization-dc\/v\/optimizing-box-volume-graphically\" target=\"_blank\" rel=\"noopener\">Watch a video about optimizing the volume of a box.<\/a><\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1165043329514\">An island is [latex]2[\/latex] mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is [latex]6[\/latex] mi west of that point. The visitor is planning to go from the cabin to the island. Suppose the visitor runs at a rate of [latex]8[\/latex] mph and swims at a rate of [latex]3[\/latex] mph. How far should the visitor run before swimming to minimize the time it takes to reach the island?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043276181\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043276181\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043276181\"><strong>Step 1:<\/strong> Let [latex]x[\/latex] be the distance running and let [latex]y[\/latex] be the distance swimming (Figure 5). Let [latex]T[\/latex] be the time it takes to get from the cabin to the island.<\/p>\n<figure style=\"width: 594px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211233\/CNX_Calc_Figure_04_07_011.jpg\" alt=\"The cabin is x miles from the shore. From that point on the shore, the island is y miles away. If you were to continue the line from the cabin to the shore (the x miles one) and if you were to draw a line from the island parallel to the shore, then the lines would extend 2 miles from the island and 6 miles from the cabin before intersecting.\" width=\"594\" height=\"352\" \/><figcaption class=\"wp-caption-text\">Figure 5. How can we choose [latex]x[\/latex] and [latex]y[\/latex] to minimize the travel time from the cabin to the island?<\/figcaption><\/figure>\n<p id=\"fs-id1165042330905\"><strong>Step 2<\/strong>: The problem is to minimize [latex]T[\/latex].<\/p>\n<p id=\"fs-id1165043346985\"><strong>Step 3<\/strong>: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent swimming. Since Distance [latex]=[\/latex] Rate [latex]\\times[\/latex] Time [latex](D=R \\times T)[\/latex], the time spent running is<\/p>\n<div id=\"fs-id1165043431452\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T_{\\text{running}}=\\frac{D_{\\text{running}}}{R_{\\text{running}}}=\\frac{x}{8}[\/latex],<\/div>\n<p>and the time spent swimming is,<\/p>\n<div id=\"fs-id1165042369650\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T_{\\text{swimming}}=\\frac{D_{\\text{swimming}}}{R_{\\text{swimming}}}=\\frac{y}{3}[\/latex]<\/div>\n<p id=\"fs-id1165043262597\">Therefore, the total time spent traveling is,<\/p>\n<div id=\"fs-id1165043433402\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=\\frac{x}{8}+\\frac{y}{3}[\/latex]<\/div>\n<p id=\"fs-id1165042517820\"><strong>Step 4<\/strong>: From Figure 5, the line segment of [latex]y[\/latex] miles forms the hypotenuse of a right triangle with legs of length [latex]2[\/latex] mi and [latex]6-x[\/latex] mi. Therefore, by the Pythagorean theorem, [latex]2^2+(6-x)^2=y^2[\/latex], and we obtain [latex]y=\\sqrt{(6-x)^2+4}[\/latex]. Thus, the total time spent traveling is given by the function<\/p>\n<div id=\"fs-id1165043430018\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T(x)=\\frac{x}{8}+\\frac{\\sqrt{(6-x)^2+4}}{3}[\/latex]<\/div>\n<p id=\"fs-id1165042328992\"><strong>Step 5<\/strong>: From Figure 5, we see that [latex]0\\le x\\le 6[\/latex]. Therefore, [latex][0,6][\/latex] is the domain of consideration.<\/p>\n<p id=\"fs-id1165043066617\"><strong>Step 6<\/strong>: Since [latex]T(x)[\/latex] is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let\u2019s begin by looking for any critical points of [latex]T[\/latex] over the interval [latex][0,6][\/latex]. The derivative is,<\/p>\n<div id=\"fs-id1165043353795\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T^{\\prime}(x)=\\frac{1}{8}-\\frac{1}{2}\\frac{[(6-x)^2+4]^{-1\/2}}{3} \\cdot 2(6-x)=\\frac{1}{8}-\\frac{(6-x)}{3\\sqrt{(6-x)^2+4}}[\/latex]<\/div>\n<p id=\"fs-id1165042516379\">If [latex]T^{\\prime}(x)=0[\/latex], then,<\/p>\n<div id=\"fs-id1165042705986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{8}=\\frac{6-x}{3\\sqrt{(6-x)^2+4}}[\/latex]<\/div>\n<p id=\"fs-id1165043107403\">Therefore,<\/p>\n<div id=\"fs-id1165042343282\" class=\"equation\" style=\"text-align: center;\">[latex]3\\sqrt{(6-x)^2+4}=8(6-x)[\/latex]<\/div>\n<p id=\"fs-id1165043284394\">Squaring both sides of this equation, we see that if [latex]x[\/latex] satisfies this equation, then [latex]x[\/latex] must satisfy,<\/p>\n<div id=\"fs-id1165043103774\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]9[(6-x)^2+4]=64(6-x)^2[\/latex],<\/div>\n<p id=\"fs-id1165042923339\">which implies,<\/p>\n<div id=\"fs-id1165043285288\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]55(6-x)^2=36[\/latex]<\/div>\n<p id=\"fs-id1165042711030\">We conclude that if [latex]x[\/latex] is a critical point, then [latex]x[\/latex] satisfies,<\/p>\n<div id=\"fs-id1165042923510\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex](x-6)^2=\\frac{36}{55}[\/latex]<\/div>\n<p id=\"fs-id1165043396085\">Therefore, the possibilities for critical points are,<\/p>\n<div id=\"fs-id1165042350053\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x=6 \\pm \\frac{6}{\\sqrt{55}}[\/latex]<\/div>\n<p id=\"fs-id1165042616899\">Since [latex]x=6+\\frac{6}{\\sqrt{55}}[\/latex] is not in the domain, it is not a possibility for a critical point. On the other hand, [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] is in the domain.<\/p>\n<p>Since we squared both sides of the equation to arrive at the possible critical points, it remains to verify that it is satisfied by [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex]. Since [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] does satisfy that equation, we conclude that [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex] is a critical point, and it is the only one.<\/p>\n<p>To justify that the time is minimized for this value of [latex]x[\/latex], we just need to check the values of [latex]T(x)[\/latex] at the endpoints [latex]x=0[\/latex] and [latex]x=6[\/latex], and compare them with the value of [latex]T(x)[\/latex] at the critical point [latex]x=6-\\frac{6}{\\sqrt{55}}[\/latex]. We find that [latex]T(0)\\approx 2.108[\/latex] h and [latex]T(6)\\approx 1.417[\/latex] h, whereas [latex]T(6-\\frac{6}{\\sqrt{55}})\\approx 1.368[\/latex] h.<\/p>\n<p>Therefore, we conclude that [latex]T[\/latex] has a local minimum at [latex]x\\approx 5.19[\/latex] mi.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>In business, companies are interested in <span class=\"no-emphasis\">maximizing revenue<\/span>. In the following example, we consider a scenario in which a company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a car. Let\u2019s use these data to determine the price the company should charge to maximize the amount of money it brings in.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1165042515832\">Owners of a car rental company have determined that if they charge customers [latex]p[\/latex] dollars per day to rent a car, where [latex]50\\le p\\le 200[\/latex], the number of cars [latex]n[\/latex] they rent per day can be modeled by the linear function [latex]n(p)=1000-5p[\/latex]. If they charge [latex]$50[\/latex] per day or less, they will rent all their cars. If they charge [latex]$200[\/latex] per day or more, they will not rent any cars. Assuming the owners plan to charge customers between [latex]$50[\/latex] per day and [latex]$200[\/latex] per day to rent a car, how much should they charge to maximize their revenue?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043423441\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043423441\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043423441\"><strong>Step 1<\/strong>: Let [latex]p[\/latex] be the price charged per car per day and let [latex]n[\/latex] be the number of cars rented per day. Let [latex]R[\/latex] be the revenue per day.<\/p>\n<p id=\"fs-id1165043431290\"><strong>Step 2<\/strong>: The problem is to maximize [latex]R[\/latex].<\/p>\n<p id=\"fs-id1165043087644\"><strong>Step 3<\/strong>: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day\u2014that is, [latex]R=n \\times p[\/latex].<\/p>\n<p id=\"fs-id1165042514435\"><strong>Step 4<\/strong>: Since the number of cars rented per day is modeled by the linear function [latex]n(p)=1000-5p[\/latex], the revenue [latex]R[\/latex] can be represented by the function:<\/p>\n<div id=\"fs-id1165043256835\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(p)=n \\times p=(1000-5p)p=-5p^2+1000p[\/latex]<\/div>\n<p id=\"fs-id1165042406610\"><strong>Step 5<\/strong>: Since the owners plan to charge between [latex]$50[\/latex] per car per day and [latex]$200[\/latex] per car per day, the problem is to find the maximum revenue [latex]R(p)[\/latex] for [latex]p[\/latex] in the closed interval [latex][50,200][\/latex].<\/p>\n<p id=\"fs-id1165043173962\"><strong>Step 6<\/strong>: Since [latex]R[\/latex] is a continuous function over the closed, bounded interval [latex][50,200][\/latex], it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is [latex]R^{\\prime}(p)=-10p+1000[\/latex]. Therefore, the critical point is [latex]p=100[\/latex].<\/p>\n<p>When [latex]p=100[\/latex], [latex]R(100)=$50,000[\/latex]. When [latex]p=50[\/latex], [latex]R(p)=$37,500[\/latex]. When [latex]p=200[\/latex], [latex]R(p)=$0[\/latex]. Therefore, the absolute maximum occurs at [latex]p=$100[\/latex].<\/p>\n<p>The car rental company should charge [latex]$100[\/latex] per day per car to maximize revenue as shown in the following figure.<\/p>\n<figure style=\"width: 495px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211236\/CNX_Calc_Figure_04_07_010.jpg\" alt=\"The function R(p) is graphed. At its maximum there is an intersection of two dashed lines and text that reads \u201cMaximum revenue is 50,000 per day when the price charged per car is 100 per day.\u201d\" width=\"495\" height=\"270\" \/><figcaption class=\"wp-caption-text\">Figure 6. To maximize revenue, a car rental company has to balance the price of a rental against the number of cars people will rent at that price.<\/figcaption><\/figure>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_lNDHOV2FlQ?controls=0&amp;start=913&amp;end=1024&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.7AppliedOptimizationProblems913to1024_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.7 Applied Optimization Problems&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1165042456921\">A rectangle is to be inscribed in the ellipse<\/p>\n<div id=\"fs-id1165042456924\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{x^2}{4}+y^2=1[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042645676\">What should the dimensions of the rectangle be to maximize its area? What is the maximum area?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042519826\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042519826\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042519826\"><strong>Step 1<\/strong>: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let [latex]L[\/latex] be the length of the rectangle and [latex]W[\/latex] be its width. Let [latex]A[\/latex] be the area of the rectangle.<\/p>\n<figure style=\"width: 575px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211239\/CNX_Calc_Figure_04_07_007.jpg\" alt=\"The ellipse x2\/4 + y2 = 1 is drawn with its x intercepts being \u00b12 and its y intercepts being \u00b11. There is a rectangle inscribed in the ellipse with length L (in the x-direction) and width W.\" width=\"575\" height=\"350\" \/><figcaption class=\"wp-caption-text\">Figure 7. We want to maximize the area of a rectangle inscribed in an ellipse.<\/figcaption><\/figure>\n<p id=\"fs-id1165042371013\"><strong>Step 2<\/strong>: The problem is to maximize [latex]A[\/latex].<\/p>\n<p id=\"fs-id1165042377481\"><strong>Step 3<\/strong>: The area of the rectangle is [latex]A=L \\times W[\/latex].<\/p>\n<p id=\"fs-id1165043257695\"><strong>Step 4<\/strong>: Let [latex](x,y)[\/latex] be the corner of the rectangle that lies in the first quadrant, as shown in Figure 7. We can write length [latex]L=2x[\/latex] and width [latex]W=2y[\/latex]. Since [latex]\\frac{x^2}{4}+y^2=1[\/latex] and [latex]y>0[\/latex], we have [latex]y=\\sqrt{1 - \\frac{x^2}{4}}[\/latex]. Therefore, the area is:<\/p>\n<div id=\"fs-id1165042320390\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=L \\times W=(2x)(2y)=4x\\sqrt{1 - \\frac{x^2}{4}}=2x\\sqrt{4-x^2}[\/latex]<\/div>\n<p id=\"fs-id1165043380076\"><strong>Step 5<\/strong>: From Figure 7, we see that to inscribe a rectangle in the ellipse, the [latex]x[\/latex]-coordinate of the corner in the first quadrant must satisfy [latex]0<x<2[\/latex]. Therefore, the problem reduces to looking for the maximum value of [latex]A(x)[\/latex] over the open interval [latex](0,2)[\/latex].<\/p>\n<p>Since [latex]A(x)[\/latex] will have an absolute maximum (and absolute minimum) over the closed interval [latex][0,2][\/latex], we consider [latex]A(x)=2x\\sqrt{4-x^2}[\/latex] over the interval [latex][0,2][\/latex]. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval.<\/p>\n<p id=\"fs-id1165042331466\"><strong>Step 6<\/strong>: As mentioned earlier, [latex]A(x)[\/latex] is a continuous function over the closed, bounded interval [latex][0,2][\/latex]. Therefore, it has an absolute maximum (and absolute minimum). At the endpoints [latex]x=0[\/latex] and [latex]x=2[\/latex], [latex]A(x)=0[\/latex]. For [latex]0<x<2[\/latex], [latex]A(x)>0[\/latex]. Therefore, the maximum must occur at a critical point. Taking the derivative of [latex]A(x)[\/latex], we obtain,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} A^{\\prime}(x) & =2\\sqrt{4-x^2}+2x \\cdot \\frac{1}{2\\sqrt{4-x^2}}(-2x) \\\\ & =2\\sqrt{4-x^2}-\\frac{2x^2}{\\sqrt{4-x^2}} \\\\ & =\\frac{8-4x^2}{\\sqrt{4-x^2}} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165043379942\">To find critical points, we need to find where [latex]A^{\\prime}(x)=0[\/latex]. We can see that if [latex]x[\/latex] is a solution of<\/p>\n<div id=\"fs-id1165043354675\" class=\"equation\" style=\"text-align: center;\">[latex]\\frac{8-4x^2}{\\sqrt{4-x^2}}=0[\/latex],<\/div>\n<p id=\"fs-id1165043183716\">then [latex]x[\/latex] must satisfy<\/p>\n<div id=\"fs-id1165043281800\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]8-4x^2=0[\/latex]<\/div>\n<p id=\"fs-id1165042515040\">Therefore, [latex]x^2=2[\/latex]. Thus, [latex]x=\\pm \\sqrt{2}[\/latex] are the possible solutions of[latex]A^{\\prime}(x)[\/latex].<\/p>\n<p>Since we are considering [latex]x[\/latex] over the interval [latex][0,2][\/latex], [latex]x=\\sqrt{2}[\/latex] is a possibility for a critical point, but [latex]x=\u2212\\sqrt{2}[\/latex] is not. Therefore, we check whether [latex]\\sqrt{2}[\/latex] is a solution of [latex]A^{\\prime}(x)[\/latex]. Since [latex]x=\\sqrt{2}[\/latex] is a solution of[latex]A^{\\prime}(x)[\/latex], we conclude that [latex]\\sqrt{2}[\/latex] is the only critical point of [latex]A(x)[\/latex] in the interval [latex][0,2][\/latex].<\/p>\n<p>Therefore, [latex]A(x)[\/latex] must have an absolute maximum at the critical point [latex]x=\\sqrt{2}[\/latex]. To determine the dimensions of the rectangle, we need to find the length [latex]L[\/latex] and the width [latex]W[\/latex]. If [latex]x=\\sqrt{2}[\/latex] then,<\/p>\n<div id=\"fs-id1165042708515\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y=\\sqrt{1-\\frac{(\\sqrt{2})^2}{4}}=\\sqrt{1-\\frac{1}{2}}=\\frac{1}{\\sqrt{2}}[\/latex]<\/div>\n<p id=\"fs-id1165043376531\">Therefore, the dimensions of the rectangle are [latex]L=2x=2\\sqrt{2}[\/latex] and [latex]W=2y=\\frac{2}{\\sqrt{2}}=\\sqrt{2}[\/latex]. The area of this rectangle is [latex]A=L \\times W=(2\\sqrt{2})(\\sqrt{2})=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288427\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288427&theme=lumen&iframe_resize_id=ohm288427&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.7 Applied Optimization Problems\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":292,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"4.7 Applied Optimization Problems","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/317"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":11,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/317\/revisions"}],"predecessor-version":[{"id":4551,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/317\/revisions\/4551"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/292"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/317\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=317"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=317"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=317"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=317"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}