{"id":655,"date":"2025-10-09T15:28:21","date_gmt":"2025-10-09T15:28:21","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=655"},"modified":"2025-10-29T13:34:03","modified_gmt":"2025-10-29T13:34:03","slug":"constrained-optimization-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/constrained-optimization-learn-it-1\/","title":{"raw":"Constrained Optimization: Learn It 1","rendered":"Constrained Optimization: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Solve optimization problems with one constraint<\/li>\r\n \t<li>Solve optimization problems with two constraints<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Lagrange Multipliers<\/h2>\r\nSolving optimization problems for functions of two or more variables can be similar to solving such problems in single-variable calculus. However, techniques for dealing with multiple variables allow us to solve more varied optimization problems for which we need to deal with additional conditions or constraints. Let's look at an example.\r\n<p class=\"whitespace-normal break-words\">Pro-T company has developed a profit model that depends on the number [latex]x[\/latex] of golf balls sold per month (measured in thousands), and the number of hours per month of advertising [latex]y[\/latex], according to the function:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]z=f(x,y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]z[\/latex] is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is [latex]50,000[\/latex], and the maximum number of hours of advertising that can be purchased is [latex]25[\/latex]. Our goal is to find the values of [latex]x[\/latex] and [latex]y[\/latex] that maximize profit.<\/p>\r\n<p class=\"whitespace-normal break-words\">Now suppose we add a budgetary constraint, such as [latex]20x+4y\\leq 216[\/latex], that takes into account the cost of producing the golf balls and the number of advertising hours purchased per month. The goal is still to maximize profit, but now there's a different type of constraint on the values of [latex]x[\/latex] and [latex]y[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">This constraint, combined with the profit function, creates what's called an <strong>optimization problem<\/strong>. The profit function [latex]f(x, y)[\/latex] is the <strong>objective function<\/strong>\u2014the function we want to maximize (or minimize in other contexts). The budgetary constraint [latex]20x+4y\\leq 216[\/latex] is the <strong>constraint function<\/strong>\u2014it limits the possible values of [latex]x[\/latex] and [latex]y[\/latex] we can consider.<\/p>\r\n<p class=\"whitespace-normal break-words\">Here's a graph of various level curves of the function [latex]f(x, y)[\/latex].<\/p>\r\n\r\n<section id=\"fs-id1167794098992\" data-depth=\"1\">[caption id=\"attachment_1292\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-1292\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23140514\/4-8-1.jpeg\" alt=\"A series of rotated ellipses that become increasingly large. The smallest one is marked f(x, y) = 400, and the biggest one is marked f(x, y) = 150.\" width=\"417\" height=\"347\" \/> Figure 1. Graph of level curves of the function\u00a0[latex]\\small{f(x, y)=48x+96y-x^{2}-2xy-9y^{2}}[\/latex] corresponding to [latex]\\small{c=150,250,350, \\text{ and } 400}[\/latex].[\/caption]<\/section><section data-depth=\"1\">\r\n<p class=\"whitespace-normal break-words\">In Figure 1, the value [latex]c[\/latex] represents different profit levels (values of the function [latex]f[\/latex]). As [latex]c[\/latex] increases, the curve shifts to the right. Since our goal is to maximize profit, we want to choose a curve as far to the right as possible. If there was no restriction on production or advertising, we could produce as many golf balls as we want and advertise as much as we want\u2014there would be no maximum profit.<\/p>\r\n<p class=\"whitespace-normal break-words\">Unfortunately, we have a budgetary constraint modeled by [latex]20x+4y\\leq 216[\/latex]. Figure 2 shows how this constraint interacts with the profit function by superimposing the line [latex]20x+4y=216[\/latex] on the previous graph.<\/p>\r\n\r\n<\/section>[caption id=\"attachment_1294\" align=\"aligncenter\" width=\"370\"]<img class=\"size-full wp-image-1294\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23140701\/4-8-2.jpeg\" alt=\"A series of rotated ellipses that become increasingly large. On the smallest ellipse, which is red, there is a tangent line marked with equation 20x + 4y = 216 that appears to touch the ellipse near (10, 4).\" width=\"370\" height=\"309\" \/> Figure 2.\u00a0Graph of level curves of the function\u00a0[latex]\\small{f(x, y)=48x+96y-x^{2}-2xy-9y^{2}}[\/latex] corresponding to [latex]\\small{c=150,250,350, \\text{ and } 395}[\/latex].\u00a0The red graph is the constraint function.[\/caption]The maximum profit occurs when the level curve is as far to the right as possible. However, the production level corresponding to this maximum must also satisfy the budgetary constraint, so the point must lie on (or to the left of) the red line in Figure 2. Inspection reveals that this point exists where the line is tangent to the level curve of [latex]f[\/latex]. Trial and error suggests this profit level is around [latex]395[\/latex], when [latex]x[\/latex] and [latex]y[\/latex] are both just less than [latex]5[\/latex].\r\n\r\nFrom a theoretical standpoint, at the point where the profit curve is tangent to the constraint line, the gradient of both functions evaluated at that point must point in the same (or opposite) direction. Recall that the gradient of a function of more than one variable is a vector. If two vectors point in the same (or opposite) directions, then one must be a constant multiple of the other. This idea is the basis of the\u00a0<strong><span id=\"dd17f45f-4704-4868-9192-90879033063c_term202\" data-type=\"term\">method of Lagrange multipliers<\/span>.<\/strong>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>method of Lagrange multipliers: one constraint<\/h3>\r\nLet\u00a0[latex]f[\/latex] and\u00a0[latex]g[\/latex]\u00a0be functions of two variables with continuous partial derivatives at every point of some open set containing the smooth curve [latex]g(x,y)=0[\/latex]. Suppose that [latex]f[\/latex], when restricted to points on the curve [latex]g(x,y)=0[\/latex], has a local extremum at the point [latex](x_0,y_0)[\/latex] and that [latex]\\nabla{g}(x_0,y_0)\\ne0[\/latex]. Then there is a number [latex]\\lambda[\/latex] called a\u00a0<strong><span id=\"dd17f45f-4704-4868-9192-90879033063c_term203\" data-type=\"term\">Lagrange multiplier<\/span><\/strong>, for which:\r\n<p style=\"text-align: center;\">[latex]\\nabla{f}(x_0,y_0)=\\lambda\\nabla{g}(x_0,y_0)[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794066211\">Assume that a constrained extremum occurs at the point [latex](x_0, y_0)[\/latex]. Furthermore, we assume that the equation [latex]g(x, y)=0[\/latex] can be smoothly parameterized as<\/p>\r\n<p style=\"text-align: center;\">[latex]x=x(s)[\/latex] and\u00a0[latex]y=y(s)[\/latex]<\/p>\r\nwhere\u00a0<em data-effect=\"italics\">s<\/em>\u00a0is an arc length parameter with reference point [latex](x_0, y_0)[\/latex] at [latex]s=0[\/latex]. Therefore, the quantity [latex]z=f(x(s), y(s))[\/latex] has a relative maximum or relative minimum at [latex]s=0[\/latex], and this implies that [latex]\\frac{dz}{ds}=0[\/latex] at that point. From the chain rule,\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{dz}{ds}=\\frac{\\partial f}{\\partial x}\\cdot\\frac{\\partial x}{\\partial s}+\\frac{\\partial f}{\\partial y}\\cdot\\frac{\\partial y}{\\partial s}=\\left(\\frac{\\partial f}{\\partial x}{\\bf{\\hat{i}}}+\\frac{\\partial f}{\\partial y}{\\bf{\\hat{j}}}\\right)\\cdot\\left(\\frac{\\partial x}{\\partial s}{\\bf{\\hat{i}}}+\\frac{\\partial y}{\\partial s}{\\bf{\\hat{j}}}\\right)=0,}[\/latex]<\/p>\r\nwhere the derivatives are all evaluated at [latex]s=0[\/latex]. However, the first factor in the dot product is the gradient of [latex]f[\/latex], and the second factor is the unit tangent vector [latex]\\text{T}(0)[\/latex] to the constraint curve. Since the point [latex](x_0, y_0)[\/latex] corresponds to [latex]s=0[\/latex], it follows from this equation that\r\n<p style=\"text-align: center;\">[latex]\\large{\\nabla{f}(x_0,y_0)\\cdot{\\text{T}}(0)=0},[\/latex]<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">which implies that the gradient is either [latex]0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">or is normal to the constraint curve at a constrained relative extremum. However, the constraint curve [latex]g(x, y)=0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is a level curve for the function [latex]g(x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">so that if [latex]\\nabla{g}(x_0,y_0)\\ne0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">then [latex]\\nabla{g}(x_0,y_0)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is normal to this curve at [latex](x_0, y_0)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">It follows, then, that there is some scalar [latex]\\lambda[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">such that<\/span>\r\n<p style=\"text-align: center;\">[latex]\\large{\\nabla{f}(x_0,y_0)=\\lambda\\nabla{g}(x_0,y_0)}[\/latex]<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">To apply the method of Lagrange multipliers to optimization problems, we need a systematic strategy.<\/p>\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"whitespace-normal break-words\"><strong>Steps for Using Lagrange Multipliers<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Identify the objective and constraint functions.<\/strong> Determine the objective function [latex]f(x, y)[\/latex] and the constraint function [latex]g(x, y)[\/latex]. Does the problem ask you to maximize or minimize?<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Set up the system of equations:<\/strong><\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\begin{array}\\nabla{f}(x_0,y_0)&amp;=\\lambda\\nabla{g}(x_0,y_0) \\\\ g(x_0,y_0)&amp;=0 \\end{array}[\/latex]<\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\" start=\"3\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Solve the system<\/strong> for [latex]x_0[\/latex] and [latex]y_0[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Evaluate and compare.<\/strong> The largest of the values of [latex]f[\/latex] at the solutions found in step 3 maximizes [latex]f[\/latex]; the smallest minimizes [latex]f[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the method of Lagrange multipliers to find the minimum value of [latex]f(x,y)=x^2+4y^2-2x+8y[\/latex] subject to the constraint [latex]x+2y=7[\/latex].[reveal-answer q=\"573132468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"573132468\"]\r\n<p id=\"fs-id1167794333406\">Let\u2019s follow the problem-solving strategy:<\/p>\r\n\r\n<ol>\r\n \t<li>The optimization function is [latex]f(x,y)=x^2+4y^2-2x+8y[\/latex]. To determine the constraint function, we must first subtract [latex]7[\/latex] from both sides of the constraint. This gives [latex]x+2y-7=0[\/latex]. The constraint function is equal to the left-hand side, so [latex]g(x,y)=x+2y-7[\/latex]. The problem asks us to solve for the minimum value of [latex]f[\/latex], subject to the constraint (see the following graph).<\/li>\r\n<\/ol>\r\n[caption id=\"attachment_1296\" align=\"aligncenter\" width=\"343\"]<img class=\"size-full wp-image-1296\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23140848\/4-8-3.jpeg\" alt=\"Two rotated ellipses, one within the other. On the largest ellipse, which is marked f(x, y) = 26, there is a tangent line marked with equation x + 2y = 7 that appears to touch the ellipse near (5, 1).\" width=\"343\" height=\"348\" \/> Figure 3.\u00a0Graph of level curves of the function\u00a0[latex]\\small{f(x, y)=x^{2}+4y^{2}-2x+8y}[\/latex] corresponding to [latex]\\small{c=10\\text{ and } 26}[\/latex].\u00a0The red graph is the constraint function.[\/caption]2. We then must calculate the gradients of both [latex]f[\/latex] and\u00a0[latex]g[\/latex]:<span data-type=\"newline\">\r\n<\/span>[latex]\\hspace{8cm}\\begin{align}\r\n\\nabla{f}(x,y)&amp;=(2x-2){\\bf{i}}+(8y+8){\\bf{j}} \\\\\r\n\\nabla{g}(x,y)&amp;={\\bf{i}}+2{\\bf{j}}.\r\n\\end{align}[\/latex]\r\nThe equation [latex]\\nabla{f}(x_0,y_0)=\\lambda\\nabla{g}(x_0,y_0)[\/latex] becomes\r\n<div id=\"fs-id1167794063017\" data-type=\"equation\" data-label=\"\">\r\n<div style=\"text-align: center;\">[latex](2x_0-2){\\bf{i}}+(8y_0+8){\\bf{j}}=\\lambda({\\bf{i}}+2{\\bf{j}}),[\/latex]<\/div>\r\n<\/div>\r\nwhich can be rewritten as\r\n<div id=\"fs-id1167794337439\" data-type=\"equation\" data-label=\"\">\r\n<div style=\"text-align: center;\">[latex](2x_0-2){\\bf{i}}+(8y_0+8){\\bf{j}}=\\lambda{\\bf{i}}+2\\lambda{\\bf{j}}.[\/latex]<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\nNext, we set the coefficients of [latex]\\bf{i}[\/latex] and [latex]\\bf{j}[\/latex] equal to each other:\r\n\r\n[latex]\\hspace{10cm}\\begin{align}\r\n\r\n2x_0-2&amp;=\\lambda \\\\\r\n\r\n8y_0+8&amp;=2\\lambda.\r\n\r\n\\end{align}[\/latex]\r\n\r\nThe equation [latex]g(x_0,y_0)=0[\/latex] becomes [latex]x_0+2y_0-7=0[\/latex]. Therefore, the system of equations that needs to be solved is\r\n\r\n[latex]\\hspace{9cm}\\begin{align}\r\n\r\n2x_0-2&amp;=\\lambda \\\\\r\n\r\n8y_0+8&amp;=2\\lambda \\\\\r\n\r\nx_0+2y_0-7&amp;=0.\r\n\r\n\\end{align}[\/latex]\r\n<div id=\"fs-id1167793964857\" data-type=\"equation\" data-label=\"\">\r\n<div><\/div>\r\n3. This is a linear system of three equations in three variables. We start by solving the second equation for [latex]\\lambda[\/latex] and substituting it into the first equation. This gives [latex]\\lambda=4y_0+4[\/latex], so substituting this into the first equation gives\r\n\r\n[latex]2x_0-2=4y_0+4[\/latex].\r\n\r\nSolving this equation for [latex]x_0[\/latex] gives [latex]x_0=2y_0+3[\/latex]. We then substitute this into the third equation:\r\n[latex]\\hspace{9cm}\\begin{align}\r\n(2y_0+3)+2y_0-7&amp;=0 \\\\\r\n4y_0-4&amp;=0 \\\\\r\ny_0=1.\r\n\\end{align}[\/latex]\r\n\r\n<span data-type=\"newline\">\r\n<\/span>Since [latex]x_0=2y_0+3[\/latex], this gives [latex]x_0=5[\/latex].\r\n\r\n4. Next, we substitute [latex](5,1)[\/latex] into [latex]f(x,y)=x^2+4y^2-2x+8y[\/latex], gives [latex]f(5,1)=5^2+4(1)^2-2(5)+8(1)=27[\/latex]. To ensure this corresponds to a minimum value on the constraint function, let\u2019s try some other values, such as the intercepts of [latex]g(x,y)=0[\/latex], Which are [latex](7,0)[\/latex] and [latex](0,3.5)[\/latex]. We get [latex]f(7,0)=35[\/latex] and\u00a0[latex]f(0.3.5)=77[\/latex],\u00a0so it appears [latex]f[\/latex] has a minimum at [latex](5,1)[\/latex].\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793299667\">Return to the golf ball manufacturer problem.\u00a0 The golf ball manufacturer, Pro-T, has developed a profit model that depends on the number [latex]x[\/latex] of golf balls sold per month (measured in thousands), and the number of hours per month of advertising [latex]y[\/latex], according to the function<\/p>\r\n<p style=\"text-align: center;\">[latex]z=f(x, y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex],<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">where [latex]z[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is measured in thousands of dollars. The budgetary constraint function relating the cost of the production of thousands golf balls and advertising units is given by [latex]20x+4y=216[\/latex].\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">Find the values of [latex]x[\/latex] and\u00a0[latex]y[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">that maximize profit, and find the maximum profit.<\/span>\r\n\r\n[reveal-answer q=\"670049531\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"670049531\"]\r\n<p id=\"fs-id1167793627947\">Again, we follow the problem-solving strategy:<\/p>\r\n\r\n<ol id=\"fs-id1167793627950\" type=\"1\">\r\n \t<li>The optimization function is [latex]f(x, y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex]. To determine the constraint function, we first subtract 216 from both sides of the constraint, then divide both sides by 4, which gives [latex]5x+y-54=0[\/latex]. The constraint function is equal to the left-hand side, so [latex]g(x, y)=5x+y-54[\/latex]. The problem asks us to solve for the maximum value of [latex]f[\/latex] subject to this constraint.<\/li>\r\n \t<li>So, we calculate the gradients of both [latex]f[\/latex] and\u00a0[latex]g[\/latex]:\r\n[latex]\\hspace{6cm}\\begin{align}\r\n\\nabla{f}(x,y)&amp;=(48-2x-2y){\\bf{i}}+(96-2x-18y){\\bf{j}} \\\\\r\n\\nabla{g}(x,y)&amp;=5{\\bf{i}}+{\\bf{j}}.\r\n\\end{align}[\/latex]\r\nThe equation [latex]\\nabla{f}(x_0,y_0)=\\lambda\\nabla{g}(x_0,y_0)[\/latex] becomes\r\n<div id=\"fs-id1167794212873\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">\r\n\r\n[latex]\\hspace{6cm}(48-2x_0-2y_0){\\bf{i}}+(96-2x_0-18y_0){\\bf{j}}=\\lambda(5{\\bf{i}}+{\\bf{j}})[\/latex],\r\nwhich can be rewritten as\r\n\r\n[latex](48-2x_0-2y_0){\\bf{i}}+(96-2x_0-18y_0){\\bf{j}}=\\lambda5{\\bf{i}}+\\lambda{\\bf{j}}[\/latex].\r\n\r\nWe then set the coefficients of [latex]\\bf{i}[\/latex] and [latex]\\bf{j}[\/latex] equal to each other:\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\n48-2x_0-2y_0&amp;=5\\lambda \\\\\r\n\r\n96-2x_0-18y_0&amp;=\\lambda.\r\n\r\n\\end{align}[\/latex]\r\n<div style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">\r\n\r\nThe equation [latex]g(x_0, y_0)=0[\/latex] becomes [latex]5x_0+y_0-54=0[\/latex]. Therefore, the system of equations that needs to be solved is\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\n48-2x_0-2y_0&amp;=5\\lambda \\\\\r\n\r\n96-2x_0-18y_0&amp;=\\lambda \\\\\r\n\r\n5x_0+y_0-54&amp;=0.\r\n\r\n\\end{align}[\/latex]\r\n\r\n<\/div>\r\n<\/div><\/li>\r\n \t<li>We use the left-hand side of the second equation to replace [latex]\\lambda[\/latex] in the first equation:\r\n<div id=\"fs-id1167793547488\" data-type=\"equation\" data-label=\"\">\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\n48-2x_0-2y_0&amp;=5(96-2x_0-18y_0) \\\\\r\n\r\n48-2x_0-2y_0&amp;=480-10x_0-90y_0 \\\\\r\n\r\n8x_0&amp;=432-88y_0 \\\\\r\n\r\nx_0&amp;=54-11y_0.\r\n\r\n\\end{align}[\/latex]\r\n\r\n<\/div>\r\nThen we substitute this into the third equation:\r\n\r\n&nbsp;\r\n<div id=\"fs-id1167794162841\" data-type=\"equation\" data-label=\"\">\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\n5(54-11y_0)+y+0&amp;=0 \\\\\r\n\r\n270-55y_0+y_0&amp;=0 \\\\\r\n\r\n216-54y_0&amp;=0 \\\\\r\n\r\ny_0&amp;=4.\r\n\r\n\\end{align}[\/latex]\r\n\r\n<\/div>\r\nSince [latex]x_0=54-11y_0[\/latex], this gives [latex]x_0=10[\/latex].<\/li>\r\n \t<li>We then substitute [latex](10, 4)[\/latex] into [latex]f(x, y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex], which gives\r\n<div id=\"fs-id1167793416705\" data-type=\"equation\" data-label=\"\">\r\n<div>\r\n\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\nf(10,4)&amp;=48(10)+96(4)-(10)^2-2(10)(4)-9(4)^2 \\\\\r\n\r\n&amp;=480+384-100-80-144=540.\r\n\r\n\\end{align}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\nTherefore the maximum profit that can be attained, subject to budgetary constraints, is [latex]$540,000[\/latex] with a production level of [latex]10,000[\/latex] golf balls and [latex]4[\/latex] hours of advertising bought per month. Let\u2019s check to make sure this truly is a maximum. The endpoints of the line that defines the constraint are [latex](10.8, 0)[\/latex] and [latex](0, 54)[\/latex]. Let\u2019s evaluate [latex]f[\/latex] at both of these points:\r\n<div id=\"fs-id1167793560949\" data-type=\"equation\" data-label=\"\">\r\n<div>\r\n\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\nf(10.8,0)&amp;=48(10.8)+96(0)-10.8^2-2(10.8)(0)-9(0)^2=401.76 \\\\\r\n\r\nf(0,54)&amp;=48(0)+96(54)-0^2-2(0)(54)-9(54)^2=-21,060.\r\n\r\n\\end{align}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\nThe second value represents a loss, since no golf balls are produced. Neither of these values exceed [latex]540[\/latex], so it seems that our extremum is a maximum value of [latex]f[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Extension to Three Variables<\/h3>\r\n<p class=\"whitespace-normal break-words\">The method of Lagrange multipliers extends naturally to optimization functions with three variables and a single constraint. For example, consider a Cobb-Douglas function [latex]f(x,y,z)=x^{0.2}y^{0.4}z^{0.4}[\/latex], where [latex]x[\/latex] represents the cost of labor, [latex]y[\/latex] represents capital input, and [latex]z[\/latex] represents the cost of advertising. The equations to solve are:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\begin{array}\\nabla{f}(x,y,z)&amp;=\\lambda\\nabla{g}(x,y,z) \\\\g(x,y,z)&amp;=0\\end{array}[\/latex]<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the minimum of the function [latex]f(x, y, z)=x^{2}+y^{2}+z^{2}[\/latex] subject to the constraint [latex]x+y+z=1[\/latex].[reveal-answer q=\"734795270\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"734795270\"]\r\n<ol id=\"fs-id1167793411204\" type=\"1\">\r\n \t<li>The optimization function is [latex]f(x, y, z)=x^{2}+y^{2}+z^{2}[\/latex]. To determine the constraint function, we subtract 1 from each side of the constraint: [latex]x+y+z-1=0[\/latex] which gives the constraint function as [latex]g(x, y, z)=x+y+z-1[\/latex].<\/li>\r\n \t<li>Next, we calculate\u00a0[latex]\\nabla{f}(x,y,z)[\/latex] and\u00a0[latex]\\nabla{g}(x,y,z)[\/latex]:\r\n<span data-type=\"newline\">[latex]\\hspace{7cm}\\begin{align}\\nabla{f}(x,y,z)&amp;=\\langle2x,2y,2z\\rangle \\\\\\nabla{g}(x,y,z)&amp;=\\langle1,1,1\\rangle \\end{align}[\/latex]<\/span><span data-type=\"newline\">\r\n<\/span>This leads to the equations<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793612617\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{7cm}\\begin{align}\\langle2x_0,2y_0,2z_0\\rangle&amp;=\\lambda\\langle1,1,1\\rangle \\\\x_0+y_0+z_0-1&amp;=0 \\end{align}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>which can be rewritten in the following form:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793373631\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n\r\n[latex]\\hspace{7cm}\\begin{align}\r\n\r\n2x_0&amp;=\\lambda \\\\\r\n\r\n2y_0&amp;=\\lambda \\\\\r\n\r\n2z_0&amp;=\\lambda \\\\\r\n\r\nx_0+y_0+z_0-1&amp;=0.\r\n\r\n\\end{align}[\/latex]\r\n\r\n<\/div><\/li>\r\n \t<li>Since each of the first three equations has [latex]\\lambda[\/latex] on the right-hand side, we know that [latex]2x_0=2y_0=2z_0[\/latex] and all three variables are equal to each other. Substituting [latex]y_0=x_0[\/latex] and [latex]z_0=x_0[\/latex] into the last equation yields [latex]3x_0-1=0[\/latex], so [latex]x_0=\\frac13[\/latex] and [latex]y_0=\\frac13[\/latex] and [latex]z_0=\\frac13[\/latex] which corresponds to a critical point on the constraint curve.<\/li>\r\n \t<li>Then, we evaluate [latex]f[\/latex] at the point [latex]\\left(\\frac13,\\frac13,\\frac13\\right)[\/latex]:\u00a0<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793937404\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\large{f\\left(\\frac13,\\frac13,\\frac13\\right)=\\left(\\frac13\\right)^2+\\left(\\frac13\\right)^2+\\left(\\frac13\\right)^2=\\frac39=\\frac13}.[\/latex]<\/div>\r\nTherefore, an extremum of the function is [latex]\\frac{1}{3}[\/latex]. To verify it is a minimum, choose other points that satisfy the constraint and calculate [latex]f[\/latex] at that point. For example,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794293267\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{7cm}\\begin{align} f(1,0,0)&amp;=1^2+0^2+0^2=1 \\\\f(0,-2,3)&amp;=0^2+(-2)^2+3^2=13. \\end{align}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Both of these values are greater than [latex]\\frac{1}{3}[\/latex], leading us to believe the extremum is a minimum.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Solve optimization problems with one constraint<\/li>\n<li>Solve optimization problems with two constraints<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Lagrange Multipliers<\/h2>\n<p>Solving optimization problems for functions of two or more variables can be similar to solving such problems in single-variable calculus. However, techniques for dealing with multiple variables allow us to solve more varied optimization problems for which we need to deal with additional conditions or constraints. Let&#8217;s look at an example.<\/p>\n<p class=\"whitespace-normal break-words\">Pro-T company has developed a profit model that depends on the number [latex]x[\/latex] of golf balls sold per month (measured in thousands), and the number of hours per month of advertising [latex]y[\/latex], according to the function:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]z=f(x,y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]z[\/latex] is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold is [latex]50,000[\/latex], and the maximum number of hours of advertising that can be purchased is [latex]25[\/latex]. Our goal is to find the values of [latex]x[\/latex] and [latex]y[\/latex] that maximize profit.<\/p>\n<p class=\"whitespace-normal break-words\">Now suppose we add a budgetary constraint, such as [latex]20x+4y\\leq 216[\/latex], that takes into account the cost of producing the golf balls and the number of advertising hours purchased per month. The goal is still to maximize profit, but now there&#8217;s a different type of constraint on the values of [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">This constraint, combined with the profit function, creates what&#8217;s called an <strong>optimization problem<\/strong>. The profit function [latex]f(x, y)[\/latex] is the <strong>objective function<\/strong>\u2014the function we want to maximize (or minimize in other contexts). The budgetary constraint [latex]20x+4y\\leq 216[\/latex] is the <strong>constraint function<\/strong>\u2014it limits the possible values of [latex]x[\/latex] and [latex]y[\/latex] we can consider.<\/p>\n<p class=\"whitespace-normal break-words\">Here&#8217;s a graph of various level curves of the function [latex]f(x, y)[\/latex].<\/p>\n<section id=\"fs-id1167794098992\" data-depth=\"1\">\n<figure id=\"attachment_1292\" aria-describedby=\"caption-attachment-1292\" style=\"width: 417px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1292\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23140514\/4-8-1.jpeg\" alt=\"A series of rotated ellipses that become increasingly large. The smallest one is marked f(x, y) = 400, and the biggest one is marked f(x, y) = 150.\" width=\"417\" height=\"347\" \/><figcaption id=\"caption-attachment-1292\" class=\"wp-caption-text\">Figure 1. Graph of level curves of the function\u00a0[latex]\\small{f(x, y)=48x+96y-x^{2}-2xy-9y^{2}}[\/latex] corresponding to [latex]\\small{c=150,250,350, \\text{ and } 400}[\/latex].<\/figcaption><\/figure>\n<\/section>\n<section data-depth=\"1\">\n<p class=\"whitespace-normal break-words\">In Figure 1, the value [latex]c[\/latex] represents different profit levels (values of the function [latex]f[\/latex]). As [latex]c[\/latex] increases, the curve shifts to the right. Since our goal is to maximize profit, we want to choose a curve as far to the right as possible. If there was no restriction on production or advertising, we could produce as many golf balls as we want and advertise as much as we want\u2014there would be no maximum profit.<\/p>\n<p class=\"whitespace-normal break-words\">Unfortunately, we have a budgetary constraint modeled by [latex]20x+4y\\leq 216[\/latex]. Figure 2 shows how this constraint interacts with the profit function by superimposing the line [latex]20x+4y=216[\/latex] on the previous graph.<\/p>\n<\/section>\n<figure id=\"attachment_1294\" aria-describedby=\"caption-attachment-1294\" style=\"width: 370px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1294\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23140701\/4-8-2.jpeg\" alt=\"A series of rotated ellipses that become increasingly large. On the smallest ellipse, which is red, there is a tangent line marked with equation 20x + 4y = 216 that appears to touch the ellipse near (10, 4).\" width=\"370\" height=\"309\" \/><figcaption id=\"caption-attachment-1294\" class=\"wp-caption-text\">Figure 2.\u00a0Graph of level curves of the function\u00a0[latex]\\small{f(x, y)=48x+96y-x^{2}-2xy-9y^{2}}[\/latex] corresponding to [latex]\\small{c=150,250,350, \\text{ and } 395}[\/latex].\u00a0The red graph is the constraint function.<\/figcaption><\/figure>\n<p>The maximum profit occurs when the level curve is as far to the right as possible. However, the production level corresponding to this maximum must also satisfy the budgetary constraint, so the point must lie on (or to the left of) the red line in Figure 2. Inspection reveals that this point exists where the line is tangent to the level curve of [latex]f[\/latex]. Trial and error suggests this profit level is around [latex]395[\/latex], when [latex]x[\/latex] and [latex]y[\/latex] are both just less than [latex]5[\/latex].<\/p>\n<p>From a theoretical standpoint, at the point where the profit curve is tangent to the constraint line, the gradient of both functions evaluated at that point must point in the same (or opposite) direction. Recall that the gradient of a function of more than one variable is a vector. If two vectors point in the same (or opposite) directions, then one must be a constant multiple of the other. This idea is the basis of the\u00a0<strong><span id=\"dd17f45f-4704-4868-9192-90879033063c_term202\" data-type=\"term\">method of Lagrange multipliers<\/span>.<\/strong><\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>method of Lagrange multipliers: one constraint<\/h3>\n<p>Let\u00a0[latex]f[\/latex] and\u00a0[latex]g[\/latex]\u00a0be functions of two variables with continuous partial derivatives at every point of some open set containing the smooth curve [latex]g(x,y)=0[\/latex]. Suppose that [latex]f[\/latex], when restricted to points on the curve [latex]g(x,y)=0[\/latex], has a local extremum at the point [latex](x_0,y_0)[\/latex] and that [latex]\\nabla{g}(x_0,y_0)\\ne0[\/latex]. Then there is a number [latex]\\lambda[\/latex] called a\u00a0<strong><span id=\"dd17f45f-4704-4868-9192-90879033063c_term203\" data-type=\"term\">Lagrange multiplier<\/span><\/strong>, for which:<\/p>\n<p style=\"text-align: center;\">[latex]\\nabla{f}(x_0,y_0)=\\lambda\\nabla{g}(x_0,y_0)[\/latex]<\/p>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\n<p id=\"fs-id1167794066211\">Assume that a constrained extremum occurs at the point [latex](x_0, y_0)[\/latex]. Furthermore, we assume that the equation [latex]g(x, y)=0[\/latex] can be smoothly parameterized as<\/p>\n<p style=\"text-align: center;\">[latex]x=x(s)[\/latex] and\u00a0[latex]y=y(s)[\/latex]<\/p>\n<p>where\u00a0<em data-effect=\"italics\">s<\/em>\u00a0is an arc length parameter with reference point [latex](x_0, y_0)[\/latex] at [latex]s=0[\/latex]. Therefore, the quantity [latex]z=f(x(s), y(s))[\/latex] has a relative maximum or relative minimum at [latex]s=0[\/latex], and this implies that [latex]\\frac{dz}{ds}=0[\/latex] at that point. From the chain rule,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{dz}{ds}=\\frac{\\partial f}{\\partial x}\\cdot\\frac{\\partial x}{\\partial s}+\\frac{\\partial f}{\\partial y}\\cdot\\frac{\\partial y}{\\partial s}=\\left(\\frac{\\partial f}{\\partial x}{\\bf{\\hat{i}}}+\\frac{\\partial f}{\\partial y}{\\bf{\\hat{j}}}\\right)\\cdot\\left(\\frac{\\partial x}{\\partial s}{\\bf{\\hat{i}}}+\\frac{\\partial y}{\\partial s}{\\bf{\\hat{j}}}\\right)=0,}[\/latex]<\/p>\n<p>where the derivatives are all evaluated at [latex]s=0[\/latex]. However, the first factor in the dot product is the gradient of [latex]f[\/latex], and the second factor is the unit tangent vector [latex]\\text{T}(0)[\/latex] to the constraint curve. Since the point [latex](x_0, y_0)[\/latex] corresponds to [latex]s=0[\/latex], it follows from this equation that<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\nabla{f}(x_0,y_0)\\cdot{\\text{T}}(0)=0},[\/latex]<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">which implies that the gradient is either [latex]0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">or is normal to the constraint curve at a constrained relative extremum. However, the constraint curve [latex]g(x, y)=0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is a level curve for the function [latex]g(x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">so that if [latex]\\nabla{g}(x_0,y_0)\\ne0[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">then [latex]\\nabla{g}(x_0,y_0)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is normal to this curve at [latex](x_0, y_0)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">It follows, then, that there is some scalar [latex]\\lambda[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">such that<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\nabla{f}(x_0,y_0)=\\lambda\\nabla{g}(x_0,y_0)}[\/latex]<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">To apply the method of Lagrange multipliers to optimization problems, we need a systematic strategy.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"whitespace-normal break-words\"><strong>Steps for Using Lagrange Multipliers<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Identify the objective and constraint functions.<\/strong> Determine the objective function [latex]f(x, y)[\/latex] and the constraint function [latex]g(x, y)[\/latex]. Does the problem ask you to maximize or minimize?<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Set up the system of equations:<\/strong><\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\begin{array}\\nabla{f}(x_0,y_0)&=\\lambda\\nabla{g}(x_0,y_0) \\\\ g(x_0,y_0)&=0 \\end{array}[\/latex]<\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\" start=\"3\">\n<li class=\"whitespace-normal break-words\"><strong>Solve the system<\/strong> for [latex]x_0[\/latex] and [latex]y_0[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Evaluate and compare.<\/strong> The largest of the values of [latex]f[\/latex] at the solutions found in step 3 maximizes [latex]f[\/latex]; the smallest minimizes [latex]f[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the method of Lagrange multipliers to find the minimum value of [latex]f(x,y)=x^2+4y^2-2x+8y[\/latex] subject to the constraint [latex]x+2y=7[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q573132468\">Show Solution<\/button><\/p>\n<div id=\"q573132468\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794333406\">Let\u2019s follow the problem-solving strategy:<\/p>\n<ol>\n<li>The optimization function is [latex]f(x,y)=x^2+4y^2-2x+8y[\/latex]. To determine the constraint function, we must first subtract [latex]7[\/latex] from both sides of the constraint. This gives [latex]x+2y-7=0[\/latex]. The constraint function is equal to the left-hand side, so [latex]g(x,y)=x+2y-7[\/latex]. The problem asks us to solve for the minimum value of [latex]f[\/latex], subject to the constraint (see the following graph).<\/li>\n<\/ol>\n<figure id=\"attachment_1296\" aria-describedby=\"caption-attachment-1296\" style=\"width: 343px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1296\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23140848\/4-8-3.jpeg\" alt=\"Two rotated ellipses, one within the other. On the largest ellipse, which is marked f(x, y) = 26, there is a tangent line marked with equation x + 2y = 7 that appears to touch the ellipse near (5, 1).\" width=\"343\" height=\"348\" \/><figcaption id=\"caption-attachment-1296\" class=\"wp-caption-text\">Figure 3.\u00a0Graph of level curves of the function\u00a0[latex]\\small{f(x, y)=x^{2}+4y^{2}-2x+8y}[\/latex] corresponding to [latex]\\small{c=10\\text{ and } 26}[\/latex].\u00a0The red graph is the constraint function.<\/figcaption><\/figure>\n<p>2. We then must calculate the gradients of both [latex]f[\/latex] and\u00a0[latex]g[\/latex]:<span data-type=\"newline\"><br \/>\n<\/span>[latex]\\hspace{8cm}\\begin{align}  \\nabla{f}(x,y)&=(2x-2){\\bf{i}}+(8y+8){\\bf{j}} \\\\  \\nabla{g}(x,y)&={\\bf{i}}+2{\\bf{j}}.  \\end{align}[\/latex]<br \/>\nThe equation [latex]\\nabla{f}(x_0,y_0)=\\lambda\\nabla{g}(x_0,y_0)[\/latex] becomes<\/p>\n<div id=\"fs-id1167794063017\" data-type=\"equation\" data-label=\"\">\n<div style=\"text-align: center;\">[latex](2x_0-2){\\bf{i}}+(8y_0+8){\\bf{j}}=\\lambda({\\bf{i}}+2{\\bf{j}}),[\/latex]<\/div>\n<\/div>\n<p>which can be rewritten as<\/p>\n<div id=\"fs-id1167794337439\" data-type=\"equation\" data-label=\"\">\n<div style=\"text-align: center;\">[latex](2x_0-2){\\bf{i}}+(8y_0+8){\\bf{j}}=\\lambda{\\bf{i}}+2\\lambda{\\bf{j}}.[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Next, we set the coefficients of [latex]\\bf{i}[\/latex] and [latex]\\bf{j}[\/latex] equal to each other:<\/p>\n<p>[latex]\\hspace{10cm}\\begin{align}    2x_0-2&=\\lambda \\\\    8y_0+8&=2\\lambda.    \\end{align}[\/latex]<\/p>\n<p>The equation [latex]g(x_0,y_0)=0[\/latex] becomes [latex]x_0+2y_0-7=0[\/latex]. Therefore, the system of equations that needs to be solved is<\/p>\n<p>[latex]\\hspace{9cm}\\begin{align}    2x_0-2&=\\lambda \\\\    8y_0+8&=2\\lambda \\\\    x_0+2y_0-7&=0.    \\end{align}[\/latex]<\/p>\n<div id=\"fs-id1167793964857\" data-type=\"equation\" data-label=\"\">\n<div><\/div>\n<p>3. This is a linear system of three equations in three variables. We start by solving the second equation for [latex]\\lambda[\/latex] and substituting it into the first equation. This gives [latex]\\lambda=4y_0+4[\/latex], so substituting this into the first equation gives<\/p>\n<p>[latex]2x_0-2=4y_0+4[\/latex].<\/p>\n<p>Solving this equation for [latex]x_0[\/latex] gives [latex]x_0=2y_0+3[\/latex]. We then substitute this into the third equation:<br \/>\n[latex]\\hspace{9cm}\\begin{align}  (2y_0+3)+2y_0-7&=0 \\\\  4y_0-4&=0 \\\\  y_0=1.  \\end{align}[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Since [latex]x_0=2y_0+3[\/latex], this gives [latex]x_0=5[\/latex].<\/p>\n<p>4. Next, we substitute [latex](5,1)[\/latex] into [latex]f(x,y)=x^2+4y^2-2x+8y[\/latex], gives [latex]f(5,1)=5^2+4(1)^2-2(5)+8(1)=27[\/latex]. To ensure this corresponds to a minimum value on the constraint function, let\u2019s try some other values, such as the intercepts of [latex]g(x,y)=0[\/latex], Which are [latex](7,0)[\/latex] and [latex](0,3.5)[\/latex]. We get [latex]f(7,0)=35[\/latex] and\u00a0[latex]f(0.3.5)=77[\/latex],\u00a0so it appears [latex]f[\/latex] has a minimum at [latex](5,1)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793299667\">Return to the golf ball manufacturer problem.\u00a0 The golf ball manufacturer, Pro-T, has developed a profit model that depends on the number [latex]x[\/latex] of golf balls sold per month (measured in thousands), and the number of hours per month of advertising [latex]y[\/latex], according to the function<\/p>\n<p style=\"text-align: center;\">[latex]z=f(x, y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex],<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">where [latex]z[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is measured in thousands of dollars. The budgetary constraint function relating the cost of the production of thousands golf balls and advertising units is given by [latex]20x+4y=216[\/latex].\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">Find the values of [latex]x[\/latex] and\u00a0[latex]y[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">that maximize profit, and find the maximum profit.<\/span><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q670049531\">Show Solution<\/button><\/p>\n<div id=\"q670049531\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793627947\">Again, we follow the problem-solving strategy:<\/p>\n<ol id=\"fs-id1167793627950\" type=\"1\">\n<li>The optimization function is [latex]f(x, y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex]. To determine the constraint function, we first subtract 216 from both sides of the constraint, then divide both sides by 4, which gives [latex]5x+y-54=0[\/latex]. The constraint function is equal to the left-hand side, so [latex]g(x, y)=5x+y-54[\/latex]. The problem asks us to solve for the maximum value of [latex]f[\/latex] subject to this constraint.<\/li>\n<li>So, we calculate the gradients of both [latex]f[\/latex] and\u00a0[latex]g[\/latex]:<br \/>\n[latex]\\hspace{6cm}\\begin{align}  \\nabla{f}(x,y)&=(48-2x-2y){\\bf{i}}+(96-2x-18y){\\bf{j}} \\\\  \\nabla{g}(x,y)&=5{\\bf{i}}+{\\bf{j}}.  \\end{align}[\/latex]<br \/>\nThe equation [latex]\\nabla{f}(x_0,y_0)=\\lambda\\nabla{g}(x_0,y_0)[\/latex] becomes<\/p>\n<div id=\"fs-id1167794212873\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">\n<p>[latex]\\hspace{6cm}(48-2x_0-2y_0){\\bf{i}}+(96-2x_0-18y_0){\\bf{j}}=\\lambda(5{\\bf{i}}+{\\bf{j}})[\/latex],<br \/>\nwhich can be rewritten as<\/p>\n<p>[latex](48-2x_0-2y_0){\\bf{i}}+(96-2x_0-18y_0){\\bf{j}}=\\lambda5{\\bf{i}}+\\lambda{\\bf{j}}[\/latex].<\/p>\n<p>We then set the coefficients of [latex]\\bf{i}[\/latex] and [latex]\\bf{j}[\/latex] equal to each other:<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    48-2x_0-2y_0&=5\\lambda \\\\    96-2x_0-18y_0&=\\lambda.    \\end{align}[\/latex]<\/p>\n<div style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">\n<p>The equation [latex]g(x_0, y_0)=0[\/latex] becomes [latex]5x_0+y_0-54=0[\/latex]. Therefore, the system of equations that needs to be solved is<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    48-2x_0-2y_0&=5\\lambda \\\\    96-2x_0-18y_0&=\\lambda \\\\    5x_0+y_0-54&=0.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/li>\n<li>We use the left-hand side of the second equation to replace [latex]\\lambda[\/latex] in the first equation:\n<div id=\"fs-id1167793547488\" data-type=\"equation\" data-label=\"\">\n<p>[latex]\\hspace{8cm}\\begin{align}    48-2x_0-2y_0&=5(96-2x_0-18y_0) \\\\    48-2x_0-2y_0&=480-10x_0-90y_0 \\\\    8x_0&=432-88y_0 \\\\    x_0&=54-11y_0.    \\end{align}[\/latex]<\/p>\n<\/div>\n<p>Then we substitute this into the third equation:<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1167794162841\" data-type=\"equation\" data-label=\"\">\n<p>[latex]\\hspace{8cm}\\begin{align}    5(54-11y_0)+y+0&=0 \\\\    270-55y_0+y_0&=0 \\\\    216-54y_0&=0 \\\\    y_0&=4.    \\end{align}[\/latex]<\/p>\n<\/div>\n<p>Since [latex]x_0=54-11y_0[\/latex], this gives [latex]x_0=10[\/latex].<\/li>\n<li>We then substitute [latex](10, 4)[\/latex] into [latex]f(x, y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex], which gives\n<div id=\"fs-id1167793416705\" data-type=\"equation\" data-label=\"\">\n<div>\n<p>[latex]\\hspace{5cm}\\begin{align}    f(10,4)&=48(10)+96(4)-(10)^2-2(10)(4)-9(4)^2 \\\\    &=480+384-100-80-144=540.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>Therefore the maximum profit that can be attained, subject to budgetary constraints, is [latex]$540,000[\/latex] with a production level of [latex]10,000[\/latex] golf balls and [latex]4[\/latex] hours of advertising bought per month. Let\u2019s check to make sure this truly is a maximum. The endpoints of the line that defines the constraint are [latex](10.8, 0)[\/latex] and [latex](0, 54)[\/latex]. Let\u2019s evaluate [latex]f[\/latex] at both of these points:<\/p>\n<div id=\"fs-id1167793560949\" data-type=\"equation\" data-label=\"\">\n<div>\n<p>[latex]\\hspace{5cm}\\begin{align}    f(10.8,0)&=48(10.8)+96(0)-10.8^2-2(10.8)(0)-9(0)^2=401.76 \\\\    f(0,54)&=48(0)+96(54)-0^2-2(0)(54)-9(54)^2=-21,060.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>The second value represents a loss, since no golf balls are produced. Neither of these values exceed [latex]540[\/latex], so it seems that our extremum is a maximum value of [latex]f[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Extension to Three Variables<\/h3>\n<p class=\"whitespace-normal break-words\">The method of Lagrange multipliers extends naturally to optimization functions with three variables and a single constraint. For example, consider a Cobb-Douglas function [latex]f(x,y,z)=x^{0.2}y^{0.4}z^{0.4}[\/latex], where [latex]x[\/latex] represents the cost of labor, [latex]y[\/latex] represents capital input, and [latex]z[\/latex] represents the cost of advertising. The equations to solve are:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\begin{array}\\nabla{f}(x,y,z)&=\\lambda\\nabla{g}(x,y,z) \\\\g(x,y,z)&=0\\end{array}[\/latex]<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Find the minimum of the function [latex]f(x, y, z)=x^{2}+y^{2}+z^{2}[\/latex] subject to the constraint [latex]x+y+z=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q734795270\">Show Solution<\/button><\/p>\n<div id=\"q734795270\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793411204\" type=\"1\">\n<li>The optimization function is [latex]f(x, y, z)=x^{2}+y^{2}+z^{2}[\/latex]. To determine the constraint function, we subtract 1 from each side of the constraint: [latex]x+y+z-1=0[\/latex] which gives the constraint function as [latex]g(x, y, z)=x+y+z-1[\/latex].<\/li>\n<li>Next, we calculate\u00a0[latex]\\nabla{f}(x,y,z)[\/latex] and\u00a0[latex]\\nabla{g}(x,y,z)[\/latex]:<br \/>\n<span data-type=\"newline\">[latex]\\hspace{7cm}\\begin{align}\\nabla{f}(x,y,z)&=\\langle2x,2y,2z\\rangle \\\\\\nabla{g}(x,y,z)&=\\langle1,1,1\\rangle \\end{align}[\/latex]<\/span><span data-type=\"newline\"><br \/>\n<\/span>This leads to the equations<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793612617\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{7cm}\\begin{align}\\langle2x_0,2y_0,2z_0\\rangle&=\\lambda\\langle1,1,1\\rangle \\\\x_0+y_0+z_0-1&=0 \\end{align}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>which can be rewritten in the following form:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793373631\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p>[latex]\\hspace{7cm}\\begin{align}    2x_0&=\\lambda \\\\    2y_0&=\\lambda \\\\    2z_0&=\\lambda \\\\    x_0+y_0+z_0-1&=0.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/li>\n<li>Since each of the first three equations has [latex]\\lambda[\/latex] on the right-hand side, we know that [latex]2x_0=2y_0=2z_0[\/latex] and all three variables are equal to each other. Substituting [latex]y_0=x_0[\/latex] and [latex]z_0=x_0[\/latex] into the last equation yields [latex]3x_0-1=0[\/latex], so [latex]x_0=\\frac13[\/latex] and [latex]y_0=\\frac13[\/latex] and [latex]z_0=\\frac13[\/latex] which corresponds to a critical point on the constraint curve.<\/li>\n<li>Then, we evaluate [latex]f[\/latex] at the point [latex]\\left(\\frac13,\\frac13,\\frac13\\right)[\/latex]:\u00a0<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793937404\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{6cm}\\large{f\\left(\\frac13,\\frac13,\\frac13\\right)=\\left(\\frac13\\right)^2+\\left(\\frac13\\right)^2+\\left(\\frac13\\right)^2=\\frac39=\\frac13}.[\/latex]<\/div>\n<p>Therefore, an extremum of the function is [latex]\\frac{1}{3}[\/latex]. To verify it is a minimum, choose other points that satisfy the constraint and calculate [latex]f[\/latex] at that point. For example,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794293267\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\hspace{7cm}\\begin{align} f(1,0,0)&=1^2+0^2+0^2=1 \\\\f(0,-2,3)&=0^2+(-2)^2+3^2=13. \\end{align}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Both of these values are greater than [latex]\\frac{1}{3}[\/latex], leading us to believe the extremum is a minimum.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-655","chapter","type-chapter","status-publish","hentry"],"part":430,"module-header":"learn_it","content_attributions":[],"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\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/655","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/655\/revisions"}],"predecessor-version":[{"id":1833,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/655\/revisions\/1833"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/430"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/655\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=655"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=655"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=655"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=655"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}