Tackle business problems to find the best ways to increase profits, minimize costs, or maximize revenue
Use optimization methods to solve problems involving geometry
Solving Optimization Problems
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.
Solving Optimization Problems over a Closed, Bounded Interval
Optimization 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.
Here’s an example of how we might approach this.
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 100 ft of wire fencing, determine the dimensions that would create a garden of maximum area. What is the maximum area?
Figure 1. We want to determine the measurements x and y that will create a garden with a maximum area using 100 ft of fencing.
Let x denote the length of the side of the garden perpendicular to the rock wall and y denote the length of the side parallel to the rock wall. Then the area of the garden is:
A=x⋅y
We want to find the maximum possible area subject to the constraint that the total fencing is 100 ft. From Figure 1, the total amount of fencing used will be 2x+y.
Therefore, the constraint equation is:
2x+y=100
Solving this equation for y, we have y=100−2x.
Thus, we can write the area as:
A(x)=x⋅(100−2x)=100x−2x2
Before trying to maximize the area function A(x)=100x−2x2, we need to determine the domain under consideration.
To construct a rectangular garden, we certainly need the lengths of both sides to be positive. Therefore, we need x>0 and y>0. Since y=100−2x, if y>0, then x<50.
Therefore, we are trying to determine the maximum value of A(x) for x over the open interval (0,50). 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.
Therefore, let’s consider the function A(x)=100x−2x2 over the closed interval [0,50]. If the maximum value occurs at an interior point, then we have found the value x in the open interval (0,50) that maximizes the area of the garden.
Therefore, we consider the following problem:
Maximize A(x)=100x−2x2 over the interval [0,50]
As mentioned earlier, since A 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, A(x)=0. Since the area is positive for all x in the open interval (0,50), the maximum must occur at a critical point. Differentiating the function A(x), we obtain,
A′(x)=100−4x
Therefore, the only critical point is x=25 (Figure 2). We conclude that the maximum area must occur when x=25. Then we have y=100−2x=100−2(25)=50.
To maximize the area of the garden, let x=25 ft and y=50 ft. The area of this garden is 1250ft2.
Figure 2. To maximize the area of the garden, we need to find the maximum value of the function A(x)=100x−2x2.
Watch the following video to see the worked solution to the example above.
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.
Determine the maximum area if we want to make the same rectangular garden as in the example above, but we have 200 ft of fencing.
We need to maximize the function A(x)=200x−2x2 over the interval [0,100].
The maximum area is 5000ft2.
Here is a general strategy for solving optimization problems similar to these above.
How To: Solve Optimization Problems
Identify All Variables: If applicable, sketch the problem scenario and label all variables.
Determine Objective: Identify which quantity needs to be maximized or minimized, and specify the range of values for any other relevant variables.
Develop a Formula: Write a formula for the objective quantity in terms of the variables, which may involve multiple variables.
Formulate Equations: Relate independent variables with any equations necessary to express the objective quantity as a function of one variable.
Set Domain: Determine the domain of consideration for the function based on the practical constraints of the problem.
Find Extremes: Calculate the maximum or minimum value of the function, typically by identifying critical points and evaluating the function at endpoints.
Let’s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be used.
An open-top box is to be made from a 24 in. by 36 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?
Step 1: Let x 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 V be the volume of the resulting box.
Figure 3. A square with side length x inches is removed from each corner of the piece of cardboard. The remaining flaps are folded to form an open-top box.
Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize V.
Step 3: As mentioned in step 2, are trying to maximize the volume of a box. The volume of a box is V=L⋅W⋅H, where L,W, and H are the length, width, and height, respectively.
Step 4: From Figure 3, we see that the height of the box is x inches, the length is 36−2x inches, and the width is 24−2x inches. Therefore, the volume of the box is,
V(x)=(36−2x)(24−2x)x=4x3−120x2+864x
Step 5: To determine the domain of consideration, let’s examine Figure 3. Certainly, we need x>0. 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 x over the open interval (0,12). Since V is a continuous function over the closed interval [0,12], we know V will have an absolute maximum over the closed interval. Therefore, we consider V over the closed interval [0,12] and check whether the absolute maximum occurs at an interior point.
Step 6: Since V(x) is a continuous function over the closed, bounded interval [0,12], V must have an absolute maximum (and an absolute minimum). Since V(x)=0 at the endpoints and V(x)>0 for [latex]0
V′(x)=12x2−240x+864
To find the critical points, we need to solve the equation:
12x2−240x+864=0
Dividing both sides of this equation by 12, the problem simplifies to solving the equation:
x2−20x+72=0
Using the quadratic formula, we find that the critical points are,
x=20±√(−20)2−4(1)(72)2=20±√1122=20±4√72=10±2√7
Since 10+2√7 is not in the domain of consideration, the only critical point we need to consider is 10−2√7. Therefore, the volume is maximized if we let x=10−2√7 in.
The maximum volume is V(10−2√7)=640+448√7≈1825in3 as shown in the following graph.
Figure 4. Maximizing the volume of the box leads to finding the maximum value of a cubic polynomial.
An island is 2 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the shore that is 6 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 8 mph and swims at a rate of 3 mph. How far should the visitor run before swimming to minimize the time it takes to reach the island?
Step 1: Let x be the distance running and let y be the distance swimming (Figure 5). Let T be the time it takes to get from the cabin to the island.
Figure 5. How can we choose x and y to minimize the travel time from the cabin to the island?
Step 2: The problem is to minimize T.
Step 3: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent swimming. Since Distance = Rate × Time (D=R×T), the time spent running is
Trunning=DrunningRrunning=x8,
and the time spent swimming is,
Tswimming=DswimmingRswimming=y3
Therefore, the total time spent traveling is,
T=x8+y3
Step 4: From Figure 5, the line segment of y miles forms the hypotenuse of a right triangle with legs of length 2 mi and 6−x mi. Therefore, by the Pythagorean theorem, 22+(6−x)2=y2, and we obtain y=√(6−x)2+4. Thus, the total time spent traveling is given by the function
T(x)=x8+√(6−x)2+43
Step 5: From Figure 5, we see that 0≤x≤6. Therefore, [0,6] is the domain of consideration.
Step 6: Since T(x) is a continuous function over a closed, bounded interval, it has a maximum and a minimum. Let’s begin by looking for any critical points of T over the interval [0,6]. The derivative is,
Squaring both sides of this equation, we see that if x satisfies this equation, then x must satisfy,
9[(6−x)2+4]=64(6−x)2,
which implies,
55(6−x)2=36
We conclude that if x is a critical point, then x satisfies,
(x−6)2=3655
Therefore, the possibilities for critical points are,
x=6±6√55
Since x=6+6√55 is not in the domain, it is not a possibility for a critical point. On the other hand, x=6−6√55 is in the domain.
Since we squared both sides of the equation to arrive at the possible critical points, it remains to verify that it is satisfied by x=6−6√55. Since x=6−6√55 does satisfy that equation, we conclude that x=6−6√55 is a critical point, and it is the only one.
To justify that the time is minimized for this value of x, we just need to check the values of T(x) at the endpoints x=0 and x=6, and compare them with the value of T(x) at the critical point x=6−6√55. We find that T(0)≈2.108 h and T(6)≈1.417 h, whereas T(6−6√55)≈1.368 h.
Therefore, we conclude that T has a local minimum at x≈5.19 mi.
In business, companies are interested in maximizing revenue. 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’s use these data to determine the price the company should charge to maximize the amount of money it brings in.
Owners of a car rental company have determined that if they charge customers p dollars per day to rent a car, where 50≤p≤200, the number of cars n they rent per day can be modeled by the linear function n(p)=1000−5p. If they charge $50 per day or less, they will rent all their cars. If they charge $200 per day or more, they will not rent any cars. Assuming the owners plan to charge customers between $50 per day and $200 per day to rent a car, how much should they charge to maximize their revenue?
Step 1: Let p be the price charged per car per day and let n be the number of cars rented per day. Let R be the revenue per day.
Step 2: The problem is to maximize R.
Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per day—that is, R=n×p.
Step 4: Since the number of cars rented per day is modeled by the linear function n(p)=1000−5p, the revenue R can be represented by the function:
R(p)=n×p=(1000−5p)p=−5p2+1000p
Step 5: Since the owners plan to charge between $50 per car per day and $200 per car per day, the problem is to find the maximum revenue R(p) for p in the closed interval [50,200].
Step 6: Since R is a continuous function over the closed, bounded interval [50,200], it has an absolute maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points. The derivative is R′(p)=−10p+1000. Therefore, the critical point is p=100.
When p=100, R(100)=$50,000. When p=50, R(p)=$37,500. When p=200, R(p)=$0. Therefore, the absolute maximum occurs at p=$100.
The car rental company should charge $100 per day per car to maximize revenue as shown in the following figure.
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.
Watch the following video to see the worked solution to this example.
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.
What should the dimensions of the rectangle be to maximize its area? What is the maximum area?
Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let L be the length of the rectangle and W be its width. Let A be the area of the rectangle.
Figure 7. We want to maximize the area of a rectangle inscribed in an ellipse.
Step 2: The problem is to maximize A.
Step 3: The area of the rectangle is A=L×W.
Step 4: Let (x,y) be the corner of the rectangle that lies in the first quadrant, as shown in Figure 7. We can write length L=2x and width W=2y. Since x24+y2=1 and y>0, we have y=√1−x24. Therefore, the area is:
A=L×W=(2x)(2y)=4x√1−x24=2x√4−x2
Step 5: From Figure 7, we see that to inscribe a rectangle in the ellipse, the x-coordinate of the corner in the first quadrant must satisfy [latex]0
Since A(x) will have an absolute maximum (and absolute minimum) over the closed interval [0,2], we consider A(x)=2x√4−x2 over the interval [0,2]. If the absolute maximum occurs at an interior point, then we have found an absolute maximum in the open interval.
Step 6: As mentioned earlier, A(x) is a continuous function over the closed, bounded interval [0,2]. Therefore, it has an absolute maximum (and absolute minimum). At the endpoints x=0 and x=2, A(x)=0. For [latex]00[/latex]. Therefore, the maximum must occur at a critical point. Taking the derivative of A(x), we obtain,
To find critical points, we need to find where A′(x)=0. We can see that if x is a solution of
8−4x2√4−x2=0,
then x must satisfy
8−4x2=0
Therefore, x2=2. Thus, x=±√2 are the possible solutions ofA′(x).
Since we are considering x over the interval [0,2], x=√2 is a possibility for a critical point, but x=−√2 is not. Therefore, we check whether √2 is a solution of A′(x). Since x=√2 is a solution ofA′(x), we conclude that √2 is the only critical point of A(x) in the interval [0,2].
Therefore, A(x) must have an absolute maximum at the critical point x=√2. To determine the dimensions of the rectangle, we need to find the length L and the width W. If x=√2 then,
y=√1−(√2)24=√1−12=1√2
Therefore, the dimensions of the rectangle are L=2x=2√2 and W=2y=2√2=√2. The area of this rectangle is A=L×W=(2√2)(√2)=4.