{"id":875,"date":"2021-09-03T17:28:44","date_gmt":"2021-09-03T17:28:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=875"},"modified":"2022-10-29T02:17:53","modified_gmt":"2022-10-29T02:17:53","slug":"lagrange-multipliers","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/lagrange-multipliers\/","title":{"raw":"Lagrange Multipliers","rendered":"Lagrange Multipliers"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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    \r\n \t
  • Use the method of Lagrange multipliers to solve optimization problems with one constraint.<\/span><\/li>\r\n \t
  • Use the method of Lagrange multipliers to solve optimization problems with two constraints.<\/span><\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n<\/div>\r\n

    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. In this section, we examine one of the more common and useful methods for solving optimization problems with constraints.<\/p>\r\n\r\n

    \r\n

    Lagrange Multipliers<\/h2>\r\n

    Chapter Opener: Profitable Golf Balls\u00a0was an applied situation involving maximizing a profit function, subject to certain\u00a0constraints<\/span><\/strong>. In that example, the constraints involved a maximum number of golf balls that could be produced and sold in [latex]1[\/latex] month [latex](x)[\/latex], and a maximum number of advertising hours that could be purchased per month [latex](y)[\/latex]. Suppose these were combined into a budgetary constraint, such as [latex]20x+4y\\leq 216[\/latex], that took 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 is a different type of constraint on the values of [latex]x[\/latex] and [latex]y[\/latex]. This constraint, when combined with the profit function [latex]f(x, y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex], is an example of an\u00a0optimization problem<\/span><\/strong>, and the function [latex]f(x, y)[\/latex] is called the\u00a0objective function<\/span><\/strong>. A graph of various level curves of the function [latex]f(x, y)[\/latex] follows.<\/p>\r\n\r\n[caption id=\"attachment_1292\" align=\"aligncenter\" width=\"417\"]\"A 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]In\u00a0Figure 1, the value [latex]c[\/latex] represents different profit levels (i.e., values of the function [latex]f[\/latex]). As the value of [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 the number of golf balls the company could produce, or the number of units of advertising available, then we could produce as many golf balls as we want, and advertise as much as we want, and there would not be a maximum profit for the company. Unfortunately, we have a budgetary constraint that is modeled by the inequality [latex]20x+4y\\leq 216[\/latex]. To see how this constraint interacts with the profit function,\u00a0Figure 2\u00a0shows the graph of the line [latex]2x+4y=216[\/latex] superimposed on the previous graph.<\/section>[caption id=\"attachment_1294\" align=\"aligncenter\" width=\"370\"]\"A 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]As mentioned previously, the maximum profit occurs when the level curve is as far to the right as possible. However, the level of production corresponding to this maximum profit must also satisfy the budgetary constraint, so the point at which this profit occurs must also lie on (or to the left of) the red line in\u00a0Figure 2. Inspection of this graph reveals that this point exists where the line is tangent to the level curve of [latex]f[\/latex]. Trial and error reveals that this profit level seems to be around [latex]395[\/latex], when [latex]x[\/latex] and [latex]y[\/latex] are both just less than [latex]5[\/latex]. We return to the solution of this problem later in this section. From a theoretical standpoint, at the point where the profit curve is tangent to the constraint line, the gradient of both of the 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\u00a0method of Lagrange multipliers<\/span>.<\/strong>\r\n

    \r\n

    Theorem: method of lagrange multipliers: one constant<\/h3>\r\n\r\n
    \r\n\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\u00a0Lagrange multiplier<\/span><\/strong>, for which\r\n

    [latex]\\nabla{f}(x_0,y_0)=\\lambda\\nabla{g}(x_0,y_0)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

    Proof<\/h2>\r\n

    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

    [latex]x=x(s)[\/latex] and\u00a0[latex]y=y(s)[\/latex]<\/p>\r\nwhere\u00a0s<\/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

    [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

    [latex]\\large{\\nabla{f}(x_0,y_0)\\cdot{\\text{T}}(0)=0},[\/latex]<\/p>\r\nwhich implies that the gradient is either [latex]0[\/latex]\u00a0<\/span>or is normal to the constraint curve at a constrained relative extremum. However, the constraint curve [latex]g(x, y)=0[\/latex]\u00a0<\/span>is a level curve for the function [latex]g(x, y)[\/latex]\u00a0<\/span>so that if [latex]\\nabla{g}(x_0,y_0)\\ne0[\/latex]\u00a0<\/span>then [latex]\\nabla{g}(x_0,y_0)[\/latex]\u00a0<\/span>is normal to this curve at [latex](x_0, y_0)[\/latex]\u00a0<\/span>It follows, then, that there is some scalar [latex]\\lambda[\/latex]\u00a0<\/span>such that<\/span>\r\n

    [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\nTo apply the\u00a0Method of Lagrange Multipliers: One Constraint\u00a0to an optimization problem similar to that for the golf ball manufacturer, we need a problem-solving strategy.\r\n

    \r\n
    \r\n
    \r\n

    Problem solving strategy: steps for using Lagrange multipliers<\/h3>\r\n\r\n
    \r\n\r\n
      \r\n \t
    1. Determine the objective function [latex]f(x, y)[\/latex] and the constraint function [latex]g(x, y)[\/latex]. Does the optimization problem involve maximizing or minimizing the objective function?<\/li>\r\n \t
    2. Set up a system of equations using the following template:\r\n

      [latex]\\hspace{8cm}\\begin{align}<\/p>\r\n\\nabla{f}(x_0,y_0)\\cdot{\\text{T}}(0)&=0 \\\\ g(x_0,y_0)&=0. \\end{align}[\/latex]<\/li>\r\n \t

    3. Solve for [latex]x_0[\/latex] and\u00a0[latex]y_0[\/latex].<\/li>\r\n \t
    4. The largest of the values of [latex]f[\/latex] at the solutions found in step 3 maximizes [latex]f[\/latex]; the smallest of those values minimizes [latex]f[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n
      \r\n

      Example: using lagrange multipliers<\/h3>\r\nUse 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].\r\n\r\n[reveal-answer q=\"573132468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"573132468\"]\r\n

      Let\u2019s follow the problem-solving strategy:<\/p>\r\n\r\n

        \r\n \t
      1. 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\"]\"Two 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]:\r\n<\/span>[latex]\\hspace{8cm}\\begin{align}\r\n\\nabla{f}(x,y)&=(2x-2){\\bf{i}}+(8y+8){\\bf{j}} \\\\\r\n\\nabla{g}(x,y)&={\\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
        \r\n
        [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
        \r\n
        [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 \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&=\\lambda \\\\\r\n\r\n8y_0+8&=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&=\\lambda \\\\\r\n\r\n8y_0+8&=2\\lambda \\\\\r\n\r\nx_0+2y_0-7&=0.\r\n\r\n\\end{align}[\/latex]\r\n
        \r\n
        <\/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&=0 \\\\\r\n4y_0-4&=0 \\\\\r\ny_0=1.\r\n\\end{align}[\/latex]\r\n\r\n\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<\/div>\r\n
        \r\n

        try it<\/h3>\r\nUse the method of Lagrange multipliers to find the maximum value of\u00a0[latex]f(x, y)=9x^{2}+36xy-4y^{2}-18x-8y[\/latex] subject to the constraint\u00a0[latex]3x+4y=32[\/latex].\r\n\r\n[reveal-answer q=\"881245602\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"881245602\"]\r\n\r\n[latex]f[\/latex] has a maximum value of [latex]976[\/latex] at the point\u00a0[latex](8, 2)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nLet\u2019s now return to the problem posed at the beginning of the section.\r\n
        \r\n

        Example: golf balls and lagrange multipliers<\/h3>\r\n

        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

        [latex]z=f(x, y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex],<\/p>\r\nwhere [latex]z[\/latex]\u00a0<\/span>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>Find the values of [latex]x[\/latex] and\u00a0[latex]y[\/latex]\u00a0<\/span>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

        Again, we follow the problem-solving strategy:<\/p>\r\n\r\n

          \r\n \t
        1. 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
        2. 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)&=(48-2x-2y){\\bf{i}}+(96-2x-18y){\\bf{j}} \\\\\r\n\\nabla{g}(x,y)&=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
          \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&=5\\lambda \\\\\r\n\r\n96-2x_0-18y_0&=\\lambda.\r\n\r\n\\end{align}[\/latex]\r\n
          \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&=5\\lambda \\\\\r\n\r\n96-2x_0-18y_0&=\\lambda \\\\\r\n\r\n5x_0+y_0-54&=0.\r\n\r\n\\end{align}[\/latex]\r\n\r\n<\/div>\r\n<\/div><\/li>\r\n \t
        3. We use the left-hand side of the second equation to replace [latex]\\lambda[\/latex] in the first equation:\r\n
          \r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\n48-2x_0-2y_0&=5(96-2x_0-18y_0) \\\\\r\n\r\n48-2x_0-2y_0&=480-10x_0-90y_0 \\\\\r\n\r\n8x_0&=432-88y_0 \\\\\r\n\r\nx_0&=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 \r\n
          \r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\n5(54-11y_0)+y+0&=0 \\\\\r\n\r\n270-55y_0+y_0&=0 \\\\\r\n\r\n216-54y_0&=0 \\\\\r\n\r\ny_0&=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
        4. We then substitute [latex](10, 4)[\/latex] into [latex]f(x, y)=48x+96y-x^{2}-2xy-9y^{2}[\/latex], which gives\r\n
          \r\n
          \r\n\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\nf(10,4)&=48(10)+96(4)-(10)^2-2(10)(4)-9(4)^2 \\\\\r\n\r\n&=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
          \r\n
          \r\n\r\n[latex]\\hspace{5cm}\\begin{align}\r\n\r\nf(10.8,0)&=48(10.8)+96(0)-10.8^2-2(10.8)(0)-9(0)^2=401.76 \\\\\r\n\r\nf(0,54)&=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<\/div>\r\n
          \r\n

          try it<\/h3>\r\nA company has determined that its production level is given by the Cobb-Douglas function [latex]f(x,y)=2.5x^{0.45}y^{0.55}[\/latex] where [latex]x[\/latex] represents the total number of labor hours in [latex]1[\/latex] year and [latex]y[\/latex] represents the total capital input for the company. Suppose\u00a0[latex]1[\/latex] unit of labor costs [latex]$40[\/latex] and\u00a0[latex]1[\/latex] unit of capital costs [latex]$50[\/latex]. Use the method of Lagrange multipliers to find the maximum value of [latex]f(x,y)=2.5x^{0.45}y^{0.55}[\/latex] subject to a budgetary constraint of [latex]$500,000[\/latex] per year.\r\n\r\n[reveal-answer q=\"625684923\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"625684923\"]\r\n\r\nA maximum production level of [latex]13,890[\/latex] occurs with [latex]5,625[\/latex] labor hours and [latex]$5,500[\/latex] of total capital input.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n