{"id":1178,"date":"2025-10-15T15:04:48","date_gmt":"2025-10-15T15:04:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1178"},"modified":"2025-10-29T13:55:47","modified_gmt":"2025-10-29T13:55:47","slug":"constrained-optimization-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/constrained-optimization-learn-it-2\/","title":{"raw":"Constrained Optimization: Learn It 2","rendered":"Constrained Optimization: Learn It 2"},"content":{"raw":"

Problems with Two Constraints<\/h2>\r\n

The method of Lagrange multipliers can be applied to problems with more than one constraint. In this case the optimization function, [latex]w[\/latex] is a function of three variables:<\/p>\r\n

[latex]\\large{w=f(x, y, z)}[\/latex]<\/p>\r\n

and it is subject to two constraints:<\/p>\r\n

[latex]g(x, y, z)=0[\/latex] and\u00a0[latex]h(x, y, z)=0[\/latex]<\/p>\r\n

There are two Lagrange multipliers, [latex]\\lambda_1[\/latex] and [latex]\\lambda_2[\/latex], and the system of equations becomes:<\/p>\r\n[latex]\\hspace{7cm}\\large{\\begin{align}\r\n\r\n\\nabla{f}(x_0,y_0,z_0)&=\\lambda_1\\nabla{g}(x_0,y_0,z_0)+\\lambda_2\\nabla{h}(x_0,y_0,z_0) \\\\\r\n\r\ng(x_0,y_0,z_0)&=0 \\\\\r\n\r\nh(x_0,y_0,z_0)&=0.\r\n\r\n\\end{align}}[\/latex]\r\n\r\n

Find the maximum and minimum values of the function\r\n

[latex]\\large{f(x, y, z)=x^{2}+y^{2}+z^{2}}[\/latex]<\/p>\r\nsubject to the constraints\u00a0[latex]z^{2}=x^{2}+y^{2}[\/latex] and\u00a0[latex]x+y-z+1=0[\/latex].\r\n\r\n[reveal-answer q=\"568214519\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"568214519\"]\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, z)=x^{2}+y^{2}+z^{2}[\/latex]. To determine the constraint functions, we first subtract [latex]z^2[\/latex] from both sides of the first constraint, which gives [latex]x^2+y^2-z^2=0[\/latex], so [latex]g(x,y,z)=x^2+y^2-z^2[\/latex]. The second constraint function is [latex]h(x,y,z)=x+y-z+1[\/latex].<\/li>\r\n \t
  2. \r\n
    \r\n\r\nWe then calculate the gradients of [latex]f[\/latex],\u00a0[latex]g[\/latex], and\u00a0[latex]h[\/latex]:\r\n\r\n \r\n\r\n[latex]\\hspace{6cm}\\large{\\begin{align}\r\n\r\n\\nabla{f}(x,y,z)&=2x{\\bf{i}}+2y{\\bf{j}}+2z{\\bf{k}} \\\\\r\n\r\n\\nabla{g}(x,y,z)&=2x{\\bf{i}}+2y{\\bf{j}}-2z{\\bf{k}} \\\\\r\n\r\n\\nabla{h}(x,y,z)&={\\bf{i}}+{\\bf{j}}-{\\bf{k}}.\r\n\r\n\\end{align}}[\/latex]\r\n\r\n\r\n<\/span>The equation [latex]\\nabla{f}(x_0,y_0,z_0)=\\lambda_1\\nabla{g}(x_0,y_0,z_0)+\\lambda_2\\nabla{h}(x_0,y_0,z_0)[\/latex] becomes\r\n<\/span>\r\n

    [latex]\\large{2x_0{\\bf{i}}+2y_0{\\bf{j}}+2z_0{\\bf{k}}=\\lambda_(2x_0{\\bf{i}}+2y_0{\\bf{j}}-2z_0{\\bf{k}})+\\lambda_2({\\bf{i}}+{\\bf{j}}-{\\bf{k}})},[\/latex]<\/p>\r\nwhich can be rewritten as\r\n

    [latex]\\large{2x_0{\\bf{i}}+2y_0{\\bf{j}}+2z_0{\\bf{k}}=(2\\lambda_1x_0+\\lambda_2){\\bf{i}}+(2\\lambda_1y_0+\\lambda_2){\\bf{j}}-(2\\lambda_1z_0+\\lambda_2){\\bf{k}}}.[\/latex]<\/p>\r\nNext, we set the coefficients of [latex]\\bf{i}[\/latex], [latex]\\bf{j}[\/latex], and [latex]\\bf{k}[\/latex] equal to each other:\r\n[latex]\\hspace{9cm}\\large{\\begin{align}\r\n2x_0&=2\\lambda_1x_0+\\lambda_2 \\\\\r\n2y_0&=2\\lambda_1y_0+\\lambda_2 \\\\\r\n2z_0&=2\\lambda_1z_0-\\lambda_2.\r\n\\end{align}}[\/latex]\r\nThe two equations that arise from the constraints are [latex]x_0^2=x_0^2+y_0^2[\/latex] and [latex]x_0+y_0-z_0+1=0[\/latex]. Combining these equations with the previous three equations gives\r\n[latex]\\hspace{7cm}\\large{\\begin{align}\r\n2x_0&=2\\lambda_1x_0+\\lambda_2 \\\\\r\n2x_0&=2\\lambda_1y_0+\\lambda_2 \\\\\r\n2x_0&=2\\lambda_1z_0-\\lambda_2 \\\\\r\n{z_0}^2&={x_0}^2+{y_0}^2 \\\\\r\nx_0+y_0-z_0+1&=0.\r\n\\end{align}}[\/latex]\r\n\r\n<\/div><\/li>\r\n \t

  3. The first three equations contain the variable [latex]\\lambda_2[\/latex]. Solving the third equation for [latex]\\lambda_2[\/latex] and replacing into the first and second equations reduces the number of equations to four:\r\n<\/span>\r\n[latex]\\hspace{7cm}\\large{\\begin{align}\r\n2x_0&=2\\lambda_1x_0-2\\lambda_1z_0-2z_0 \\\\2y_0&=2\\lambda_1y_0-2\\lambda_1z_0-2z_0 \\\\{z_0}^2&={x_0}^2+{y_0}^2 \\\\x_0+y_0-z_0+1&=0.\\end{align}}[\/latex]\r\n\r\n<\/span>Next, we solve the first and second equation for [latex]\\lambda_1[\/latex]. The first equation gives [latex]\\lambda_1=\\frac{x_0+z_0}{x_0-z_0}[\/latex], the second equation gives [latex]\\lambda_1=\\frac{y_0+z_0}{y_0-z_0}[\/latex]. We set the right-hand side of each equation equal to each other and cross-multiply:\r\n<\/span>\r\n
    \r\n\r\n[latex]\\hspace{6cm}\\large{\\begin{align}\\frac{x_0+z_0}{x_0-z_0}&=\\frac{y_0+z_0}{y_0-z_0} \\\\\r\n\r\n(x_0+z_0)(y_0-z_0)&=(x_0-z_0)(y_0+z_0) \\\\\r\n\r\nx_0y_0-x_0z_0+y_0z_0-{z_0}^2&=x_0y_0+x_0z_0-y_0z_0-{z_0}^2 \\\\\r\n\r\n2y_0z_0-2x_0z_0&=0 \\\\\r\n\r\n2x_0(y_0-x_0)&=0.\r\n\r\n\\end{align}}[\/latex]\r\n\r\n<\/div>\r\nTherefore, either [latex]z_0=0[\/latex] or [latex]y_0=x_0[\/latex]. If [latex]z_0=0[\/latex], then the first constraint becomes [latex]0=x_0^2+y_0^2[\/latex] The only real solution to this equation is [latex]x_0=0[\/latex] and [latex]y_0=0[\/latex], which gives the ordered triple [latex](0,0,0)[\/latex]. This point does not satisfy the second constraint, so it is not a solution.Next, we consider [latex]y_0=x_0[\/latex], which reduces the number of equations to three:\r\n<\/span>\r\n\r\n[latex]\\hspace{8cm}\\large{\\begin{align}\r\n\r\ny_0&=x_0 \\\\\r\n\r\n{z_0}^2&={x_0}^2+{y_0}^2 \\\\\r\n\r\nx_0+y_0-z_0+1&=0.\r\n\r\n\\end{align}}[\/latex]\r\nWe substitute the first equation into the second and third equations:\r\n<\/span>\r\n\r\n[latex]\\hspace{8cm}\\large{\\begin{align}\r\n\r\n{z_0}^2&={x_0}^2+{y_0}^2 \\\\\r\n\r\nx_0+y_0-z_0+1&=0.\r\n\r\n\\end{align}}[\/latex]\r\n\r\nThen, we solve the second equation for [latex]z_0[\/latex], which gives [latex]z_0=2x_0+1[\/latex]. We then substitute this into the first equation,\r\n<\/span>\r\n\r\n[latex]\\hspace{7cm}\\large{\\begin{align}\r\n\r\n{z_0}^2&={2x_0}^2 \\\\\r\n\r\n(2x_0+1)^2&={2x_0}^2 \\\\\r\n\r\n{4x_0}^2+4x_0+1&={2x_0}^2 \\\\\r\n\r\n{2x_0}^2+4x_0+1&=0\r\n\r\n\\end{align}}[\/latex]\r\nand use the quadratic formula to solve for [latex]x_0[\/latex]:\r\n

    [latex]\\large{x+0=\\frac{-4\\pm\\sqrt{4^2-4(2)(1)}}{2(2)}=\\frac{-4\\pm\\sqrt8}4=\\frac{-4\\pm2\\sqrt2}4=-1\\pm\\frac{\\sqrt2}2}[\/latex].<\/span><\/p>\r\n\r\n<\/span>Recall [latex]y_0=x_0[\/latex],so this solves for [latex]y_0[\/latex] as well. Then, [latex]z_0=2x_0+1[\/latex], so\r\n

    [latex]\\large{x_0=2x_0+1=2\\left(-1\\pm\\frac{\\sqrt2}2\\right)+1=-2+1\\pm\\sqrt2=-1\\pm\\sqrt2}.[\/latex]<\/span><\/p>\r\nTherefore, there are two ordered triplet solutions:\r\n

    [latex]\\left(-1+\\frac{\\sqrt2}2,-1+\\frac{\\sqrt2}2,-1+\\sqrt2\\right)[\/latex] and [latex]\\left(-1-\\frac{\\sqrt2}2,-1-\\frac{\\sqrt2}2,-1-\\sqrt2\\right).[\/latex]<\/span><\/p>\r\n<\/li>\r\n \t

  4. We substitute [latex]\\left(-1+\\frac{\\sqrt2}2,-1+\\frac{\\sqrt2}2,-1+\\sqrt2\\right)[\/latex] into [latex]f(x,y,z)=x^2+y^2+z^2[\/latex], which gives\r\n<\/span>\r\n[latex]\\hspace{3cm}\\begin{align}\r\nf\\left(-1+\\frac{\\sqrt2}2,-1+\\frac{\\sqrt2}2,-1+\\sqrt2\\right)&=\\left(-1+\\frac{\\sqrt2}2\\right)^2+\\left(-1+\\frac{\\sqrt2}2\\right)^2+\\left(-1+\\sqrt2\\right)^2 \\\\\r\n&=(1-\\sqrt2+\\frac12)+(1-\\sqrt2+\\frac12)+(1-2\\sqrt2+2) \\\\\r\n&=6-4\\sqrt2.\r\n\\end{align}[\/latex]\r\n\r\n<\/span>[latex]6+4\\sqrt2[\/latex] is the maximum value and [latex]6-4\\sqrt2[\/latex]\u00a0is the minimum value of [latex]f(x,y,z)[\/latex] subject to the given constraints.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

    Problems with Two Constraints<\/h2>\n

    The method of Lagrange multipliers can be applied to problems with more than one constraint. In this case the optimization function, [latex]w[\/latex] is a function of three variables:<\/p>\n

    [latex]\\large{w=f(x, y, z)}[\/latex]<\/p>\n

    and it is subject to two constraints:<\/p>\n

    [latex]g(x, y, z)=0[\/latex] and\u00a0[latex]h(x, y, z)=0[\/latex]<\/p>\n

    There are two Lagrange multipliers, [latex]\\lambda_1[\/latex] and [latex]\\lambda_2[\/latex], and the system of equations becomes:<\/p>\n

    [latex]\\hspace{7cm}\\large{\\begin{align} \\nabla{f}(x_0,y_0,z_0)&=\\lambda_1\\nabla{g}(x_0,y_0,z_0)+\\lambda_2\\nabla{h}(x_0,y_0,z_0) \\\\ g(x_0,y_0,z_0)&=0 \\\\ h(x_0,y_0,z_0)&=0. \\end{align}}[\/latex]<\/p>\n

    Find the maximum and minimum values of the function<\/p>\n

    [latex]\\large{f(x, y, z)=x^{2}+y^{2}+z^{2}}[\/latex]<\/p>\n

    subject to the constraints\u00a0[latex]z^{2}=x^{2}+y^{2}[\/latex] and\u00a0[latex]x+y-z+1=0[\/latex].<\/p>\n