{"id":1167,"date":"2025-10-15T14:54:03","date_gmt":"2025-10-15T14:54:03","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1167"},"modified":"2025-10-29T12:30:14","modified_gmt":"2025-10-29T12:30:14","slug":"critical-points-and-local-extrema-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/critical-points-and-local-extrema-learn-it-3\/","title":{"raw":"Critical Points and Local Extrema: Learn It 3","rendered":"Critical Points and Local Extrema: Learn It 3"},"content":{"raw":"<h2>Absolute Maxima and Minima<\/h2>\r\nWhen finding global extrema of single-variable functions on a closed interval, we check critical values over that interval and then evaluate the function at the endpoints. Let's briefly recall how this works.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Extreme Value Theorem (Single-Variable)<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A continuous function [latex]f(x)[\/latex] on a closed interval [latex][a,b][\/latex] attains an absolute maximum value at some point and an absolute minimum value at some point. These values are the largest and smallest values found among:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The values of [latex]f[\/latex] at the critical values within [latex][a,b][\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The values of [latex]f[\/latex] at the endpoints of [latex][a,b][\/latex]<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">For functions of two variables, the closed interval is replaced by a closed, bounded set. A set is <strong>bounded<\/strong> if all points in that set can be contained within a ball (or disk) of finite radius. We need to find the critical points inside the set and calculate the corresponding critical values. Then, we find the maximum and minimum values on the boundary of the set. The largest function value corresponds to the global maximum and the smallest to the absolute minimum.<\/p>\r\n<p class=\"whitespace-normal break-words\">First, we need to be assured that such values exist.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>extreme value theorem<\/h3>\r\nA continuous function [latex]f(x, y)[\/latex] on a closed and bounded set [latex]D[\/latex] in the plane attains an absolute maximum value at some point of [latex]D[\/latex] and an absolute minimum value at some point of [latex]D[\/latex].\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Now that we know continuous functions on closed, bounded sets attain their extreme values, we need to know how to find them.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>finding absolute extrema<\/h3>\r\n<p class=\"whitespace-normal break-words\">Assume [latex]z=f(x, y)[\/latex] is a differentiable function of two variables defined on a closed, bounded set [latex]D[\/latex]. Then [latex]f[\/latex] will attain the absolute maximum value and the absolute minimum value, which are the largest and smallest values found among:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The values of [latex]f[\/latex] at the critical points of [latex]f[\/latex] in [latex]D[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The values of [latex]f[\/latex] on the boundary of [latex]D[\/latex]<\/li>\r\n<\/ul>\r\n<ol id=\"fs-id1167793417260\" type=\"i\"><\/ol>\r\n<\/section>The proof follows directly from the extreme value theorem and Fermat's theorem. If either extremum is not on the boundary of [latex]D[\/latex], then it's at an interior point of [latex]D[\/latex]. But an interior point [latex](x_0, y_0)[\/latex] that's an absolute extremum is also a local extremum, so [latex](x_0, y_0)[\/latex] is a critical point by Fermat's theorem. Therefore, the only possible values for global extrema are the extreme values on the interior or boundary of [latex]D[\/latex].\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Absolute Extrema on a Closed, Bounded Set<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Let [latex]z=f(x, y)[\/latex] be a continuous function defined on a closed, bounded set [latex]D[\/latex], and assume [latex]f[\/latex] is differentiable on [latex]D[\/latex]. To find the absolute maximum and minimum values:<\/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>Find the critical points<\/strong> of [latex]f[\/latex] in [latex]D[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Calculate<\/strong> [latex]f[\/latex] at each critical point<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Determine the maximum and minimum values<\/strong> of [latex]f[\/latex] on the boundary of [latex]D[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Compare all values<\/strong> from steps 2 and 3\u2014the maximum and minimum will occur at one of these values<\/li>\r\n<\/ol>\r\n<\/section>Finding extrema on the boundary can be challenging. If the boundary is a rectangle or set of straight lines, parameterize the line segments and determine the maxima on each segment. The same approach works for circles and ellipses.\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">If the boundary is a more complicated curve defined by [latex]g(x, y)=c[\/latex], and the first-order partial derivatives of [latex]g[\/latex] exist, the method of Lagrange multipliers can help determine extrema on the boundary.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:\r\n<div id=\"fs-id1167793637905\" data-type=\"problem\">\r\n<div class=\"os-problem-container\">\r\n<ol id=\"fs-id1167793637915\" type=\"a\">\r\n \t<li>[latex]f(x, y)=x^{2}-2xy+4y^{2}-4x-2y+24[\/latex] on the domain defined by\u00a0[latex]0\\leq x\\leq 4[\/latex] and\u00a0[latex]0\\leq y\\leq 2[\/latex].<\/li>\r\n \t<li>[latex]g(x, y)=x^{2}+y^{2}+4x-6y[\/latex] on the domain defined by\u00a0[latex]x^{2}+y^{2}\\leq 16[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"147756901\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"147756901\"]\r\n\r\na. Using the problem-solving strategy, step 1 involves finding the critical points of [latex]f[\/latex] on its domain. Therefore, we first calculate [latex]f_x(x, y)[\/latex] and [latex]f_y(x, y)[\/latex], then set them each equal to zero:\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\nf_x(x, y)&amp;=2x-2y-4 \\\\\r\n\r\nf_y(x, y)&amp;=-2x+8y-2.\r\n\r\n\\end{align}[\/latex]\r\n\r\nSetting them equal to zero yields the system of equations\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\n2x-2y-4&amp;=0 \\\\\r\n\r\n-2x+8y-2&amp;=0.\r\n\r\n\\end{align}[\/latex]\r\n\r\nThe solution to this system is [latex]x=3[\/latex] and [latex]y=1[\/latex]. Therefore [latex](3, 1)[\/latex] is a critical point of [latex]f[\/latex]. Calculating [latex]f(3, 1)[\/latex] gives [latex]f(3, 1)=17[\/latex].\u00a0<span data-type=\"newline\">\r\n<\/span>The next step involves finding the extrema of [latex]f[\/latex] on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:\r\n\r\n[caption id=\"attachment_1286\" align=\"aligncenter\" width=\"267\"]<img class=\"size-full wp-image-1286\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135426\/4-7-7.jpeg\" alt=\"A rectangle is drawn in the first quadrant with one corner at the origin, horizontal length 4, and height 2. This rectangle is marked D, and the sides are marked in counterclockwise order from the side overlapping the x axis L1, L2, L3, and L4.\" width=\"267\" height=\"272\" \/> Figure 1.\u00a0Graph of the domain of the function [latex]\\small{f(x,y)=x^{2}-2xy+4y^{2}-4x-2y+24}[\/latex].[\/caption]\r\n<div class=\"os-problem-container\">\r\n\r\n&nbsp;\r\n\r\n[latex]L_1[\/latex] is the line segment connecting\u00a0[latex](0, 0)[\/latex] and\u00a0[latex](4, 0)[\/latex], and it can be parameterized by the equations\u00a0[latex]x(t)=t[\/latex],\u00a0[latex]y(t)=0[\/latex] for\u00a0[latex]0\\leq t\\leq 4[\/latex]. Define\u00a0[latex]g(t)=f(x(t), y(t))[\/latex]. This gives\u00a0[latex]g(t)=t^{2}-4t+24[\/latex]. Differentiating\u00a0[latex]g[\/latex] leads to\u00a0[latex]g'(t)=2t-4[\/latex]. Therefore,\u00a0[latex]g[\/latex] has a critical value at\u00a0[latex]t=2[\/latex], which corresponds to the point\u00a0[latex](2, 0)[\/latex]. Calculating\u00a0[latex]f(2, 0)[\/latex] gives the [latex]z[\/latex]-value\u00a0[latex]20[\/latex].\r\n\r\n[latex]L_2[\/latex] is the line segment connecting\u00a0[latex](4, 0)[\/latex] and\u00a0[latex](4, 2)[\/latex], and it can be parameterized by the equations\u00a0[latex]x(t)=4[\/latex],\u00a0[latex]y(t)=t[\/latex] for\u00a0[latex]0\\leq t\\leq 2[\/latex]. Again, define\u00a0[latex]g(t)=f(x(t), y(t))[\/latex]. This gives\u00a0[latex]g(t)=4t^{2}-10t+24[\/latex]. Then,\u00a0[latex]g'(t)=8t-10[\/latex].\u00a0[latex]g[\/latex] has a critical value at\u00a0[latex]t=\\frac{5}{4}[\/latex], which corresponds to the point\u00a0[latex](4, \\frac{5}{4})[\/latex]. Calculating\u00a0[latex]f(4, \\frac{5}{4})[\/latex] gives the [latex]z[\/latex]-value\u00a0[latex]17.75[\/latex].\r\n\r\n[latex]L_3[\/latex] is the line segment connecting\u00a0[latex](0, 2)[\/latex] and\u00a0[latex](4, 2)[\/latex], and it can be parameterized by the equations\u00a0[latex]x(t)=t[\/latex],\u00a0[latex]y(t)=2[\/latex] for\u00a0[latex]0\\leq t\\leq 4[\/latex]. Again, define\u00a0[latex]g(t)=f(x(t), y(t))[\/latex]. This gives\u00a0[latex]g(t)=t^{2}-8t+36[\/latex]. The critical value corresponds to the point\u00a0[latex](4, 2)[\/latex]. So, calculating\u00a0[latex]f(4, 2)[\/latex] gives the [latex]z[\/latex]-value\u00a0[latex]20[\/latex].\r\n\r\n[latex]L_4[\/latex] is the line segment connecting\u00a0[latex](0, 0)[\/latex] and\u00a0[latex](0, 2)[\/latex], and it can be parameterized by the equations\u00a0[latex]x(t)=0[\/latex],\u00a0[latex]y(t)=t[\/latex] for\u00a0[latex]0\\leq t\\leq 2[\/latex]. This time [latex]g(t)=4t^{2}-2t+24[\/latex] and the critical value\u00a0[latex]t=\\frac{1}{4}[\/latex]corresponds to the point\u00a0[latex](0, \\frac{1}{4})[\/latex]. Calculating\u00a0[latex]f(0, \\frac{1}{4})[\/latex] gives the [latex]z[\/latex]-value\u00a0[latex]23.75[\/latex]. We also need to find the values of [latex]f(x, y)[\/latex] at the corners of its domain. These corners are located at [latex](0, 0)[\/latex],\u00a0[latex](4, 0)[\/latex],\u00a0[latex](4, 2)[\/latex], and\u00a0[latex](0, 2)[\/latex]:\r\n[latex]\\hspace{5cm}\\begin{align}\r\nf(0,0)&amp;=(0)^2-2(0)(0)+4(0)^2-4(0)-2(0)+24=24 \\\\\r\nf(4,0)&amp;=(4)^2-2(4)(0)+4(0)^2-4(4)-2(0)+24=24 \\\\\r\nf(4,2)&amp;=(4)^2-2(4)(2)+4(2)^2-4(4)-2(2)+24=20 \\\\\r\nf(0,2)&amp;=(0)^2-2(0)(2)+4(2)^2-4(0)-2(2)+24=36.\r\n\\end{align}[\/latex]\r\n\r\nThe absolute maximum value is [latex]36[\/latex], which occurs at [latex](0, 2)[\/latex], and the global minimum value is [latex]17[\/latex], which occurs at [latex](3, 1)[\/latex] as shown in the following figure.\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1287\" align=\"aligncenter\" width=\"729\"]<img class=\"size-full wp-image-1287\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135537\/4-7-8.jpeg\" alt=\"The function f(x, y) = x2 \u2212 2xy \u2013 4x + 4y2 \u2013 2y + 24 is shown with local minima at (4, 2, 20) and (2, 0, 20) and local maximum at (0, 2, 36). The shape is a plane curving up on the corners significantly near (0, 2) and slightly less near (4, 0).\" width=\"729\" height=\"587\" \/> Figure 2. The function [latex]\\small{f(x,y)}[\/latex] has one global minimum and one global maximum over its domain.[\/caption]b.\u00a0Using the problem-solving strategy, step 1 involves finding the critical points of [latex]g[\/latex] on its domain. Therefore, we first calculate [latex]g_x(x, y)[\/latex] and [latex]g_y(x, y)[\/latex], then set them each equal to zero:<\/div>\r\n<div><\/div>\r\n<div>\r\n\r\n[latex]\\hspace{8cm}\\begin{align}g_x(x, y)&amp;=2y+4 \\\\g_y(x, y)&amp;=2y-6.\r\n\r\n\\end{align}[\/latex]\r\n\r\nSetting them equal to zero yields the system of equations\r\n\r\n<\/div>\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\n2x+4&amp;=0 \\\\\r\n2y-6&amp;=0.\r\n\\end{align}[\/latex]\r\nThe solution to this system is [latex]x=-2[\/latex] and [latex]y=3[\/latex]. Therefore, [latex](-2, 3)[\/latex] is a critical point of [latex]g[\/latex]. Calculating [latex]g(-2, 3)[\/latex], we get\r\n<p style=\"text-align: center;\">[latex]g(-2,3)=(-2)^2+3^2+4(-2)-6(3)=4+9-8-18=-13.[\/latex]<\/p>\r\nThe next step involves finding the extrema of [latex]g[\/latex]\u00a0on the boundary of its domain. The boundary of its domain consists of a circle of radius [latex]4[\/latex] centered at the origin as shown in the following graph.\r\n\r\n[caption id=\"attachment_1288\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-1288\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135649\/4-7-9.jpeg\" alt=\"A filled-in circle marked D of radius four with center at the origin.\" width=\"417\" height=\"423\" \/> Figure 3.\u00a0Graph of the domain of the function [latex]\\small{g(x,y)=x^{2}+y^{2}+4x-6y}[\/latex].[\/caption]\r\nThe boundary of the domain of [latex]g[\/latex] can be parameterized using the functions\u00a0[latex]x(t)=4\\cos t[\/latex],\u00a0[latex]y(t)=4\\sin t[\/latex] for\u00a0[latex]0\\leq{t}\\leq2\\pi[\/latex]. Define\u00a0[latex]h(t)=g(x(t), y(t))[\/latex]:\r\n[latex]\\hspace{5cm}\\begin{align}\r\nh(t)&amp;=g(x(t),y(t)) \\\\\r\n&amp;=(4\\cos{t})^2+(4\\sin t)^2+4(4\\cos{t})-6(4\\sin t) \\\\\r\n&amp;=16\\cos^2{t}+16\\sin^2{t}+16\\cos{t}-24\\sin t \\\\\r\n&amp;=16+16\\cos{t}-24\\sin t.\\end{align}[\/latex]\r\nSetting\u00a0[latex]h'(t)=0[\/latex] leads to\r\n[latex]\\hspace{5cm}\\begin{align}-16\\sin{t}-24\\cos{t}&amp;=0 \\\\-16\\sin{t}&amp;=24\\cos{t} \\\\\\frac{-16\\sin{t}}{-16\\cos{t}}&amp;=\\frac{24\\cos{t}}{-16\\cos{t}} \\\\\\tan{t}&amp;=-\\frac32\\end{align}[\/latex]\r\nThis equation has two solutions over the interval\u00a0[latex]0\\leq{t}\\leq2\\pi[\/latex]. One is\u00a0[latex]t=\\pi-\\arctan{(\\frac32)}[\/latex] and the other is\u00a0[latex]t=2\\pi-\\arctan{(\\frac32)}[\/latex]. For the first angle,\r\n[latex]\\hspace{4cm}\\begin{align}\\sin{t}&amp;=\\sin{(\\pi-\\arctan{(\\frac32)})}=\\sin{(\\arctan{(\\frac32)})}=\\frac{3\\sqrt{13}}{13} \\\\\\cos{t}&amp;=\\cos{(\\pi-\\arctan{(\\frac32)})}=-\\cos{(\\arctan{(\\frac32)})}=-\\frac{2\\sqrt{13}}{13}\\end{align}[\/latex]\r\nTherefore, [latex]x(t)=4\\cos{t}=\\frac{8\\sqrt{13}}{13}[\/latex] and [latex]y(t)=4\\sin{t}=-\\frac{12\\sqrt{13}}{13}[\/latex], so [latex]\\left(\\frac{8\\sqrt{13}}{13},-\\frac{12\\sqrt{13}}{13}\\right)[\/latex] is a critical point on the boundary and[latex]\\hspace{4cm}\\begin{align}g\\left(-\\frac{8\\sqrt{13}}{13},\\frac{12\\sqrt{13}}{13}\\right)&amp;=\\left(-\\frac{8\\sqrt{13}}{13}\\right)^2+\\left(\\frac{12\\sqrt{13}}{13}\\right)^2+4\\left(-\\frac{8\\sqrt{13}}{13}\\right)-6\\left(\\frac{12\\sqrt{13}}{13}\\right) \\\\&amp;=\\frac{144}{13}+\\frac{64}{13}-\\frac{32\\sqrt{13}}{13}-\\frac{72\\sqrt{13}}{13} \\\\&amp;=\\frac{208-104\\sqrt{13}}{13}\\approx-12.844.\\end{align}[\/latex]\r\nFor the second angle,\r\n[latex]\\hspace{5cm}\\begin{align}\\sin{t}&amp;=\\sin{(2\\pi-\\arctan{(\\frac32)})}=-\\sin{(\\arctan{(\\frac32)})}=-\\frac{3\\sqrt{13}}{13} \\\\\\cos{t}&amp;=\\cos{(2\\pi-\\arctan{(\\frac32)})}=\\cos{(\\arctan{(\\frac32)})}=\\frac{2\\sqrt{13}}{13}.\\end{align}[\/latex]\r\nTherefore,\u00a0[latex]x(t)=4\\cos{t}=\\frac{8\\sqrt{13}}{13}[\/latex] and\u00a0[latex]y(t)=4\\sin{t}=-\\frac{12\\sqrt{13}}{13}[\/latex], so\u00a0[latex]\\left(\\frac{8\\sqrt{13}}{13},-\\frac{12\\sqrt{13}}{13}\\right)[\/latex] is a critical point on the boundary and[latex]\\hspace{4cm}\\begin{align}g\\left(\\frac{8\\sqrt{13}}{13},-\\frac{12\\sqrt{13}}{13}\\right)&amp;=\\left(\\frac{8\\sqrt{13}}{13}\\right)^2+\\left(-\\frac{12\\sqrt{13}}{13}\\right)^2+4\\left(\\frac{8\\sqrt{13}}{13}\\right)-6\\left(-\\frac{12\\sqrt{13}}{13}\\right) \\\\&amp;=\\frac{144}{13}+\\frac{64}{13}+\\frac{32\\sqrt{13}}{13}+\\frac{72\\sqrt{13}}{13} \\\\&amp;=\\frac{208+104\\sqrt{13}}{13}\\approx44.844.\\end{align}[\/latex]The absolute minimum of [latex]g[\/latex] is [latex]-13[\/latex] which is attained at the point [latex](-2, 3)[\/latex], which is an interior point of [latex]D[\/latex]. The absolute maximum of [latex]g[\/latex] is approximately equal to [latex]44.844[\/latex], which is attained at the boundary point [latex]\\left(\\frac{8\\sqrt{13}}{13},-\\frac{12\\sqrt{13}}{13}\\right)[\/latex]. These are the absolute extrema of [latex]g[\/latex] on [latex]D[\/latex] as shown in the following figure.[caption id=\"attachment_1289\" align=\"aligncenter\" width=\"688\"]<img class=\"size-full wp-image-1289\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135817\/4-7-10.jpeg\" alt=\"The function f(x, y) = x2 + 4x + y2 \u2013 6y is shown with local minimum at (\u201312\/5, 16\/5, \u201364\/5) and local maximum at (12\/5, \u221216\/5, 224\/5). The shape is a plane curving up from near (\u22124, 4) to (4, \u22124).\" width=\"688\" height=\"470\" \/> Figure 4. The function [latex]\\small{f(x,y)}[\/latex] has a local minimum and a local maximum.[\/caption]<\/div>\r\n<div class=\"os-problem-container\"><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>","rendered":"<h2>Absolute Maxima and Minima<\/h2>\n<p>When finding global extrema of single-variable functions on a closed interval, we check critical values over that interval and then evaluate the function at the endpoints. Let&#8217;s briefly recall how this works.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Extreme Value Theorem (Single-Variable)<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A continuous function [latex]f(x)[\/latex] on a closed interval [latex][a,b][\/latex] attains an absolute maximum value at some point and an absolute minimum value at some point. These values are the largest and smallest values found among:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The values of [latex]f[\/latex] at the critical values within [latex][a,b][\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">The values of [latex]f[\/latex] at the endpoints of [latex][a,b][\/latex]<\/li>\n<\/ul>\n<\/section>\n<p class=\"whitespace-normal break-words\">For functions of two variables, the closed interval is replaced by a closed, bounded set. A set is <strong>bounded<\/strong> if all points in that set can be contained within a ball (or disk) of finite radius. We need to find the critical points inside the set and calculate the corresponding critical values. Then, we find the maximum and minimum values on the boundary of the set. The largest function value corresponds to the global maximum and the smallest to the absolute minimum.<\/p>\n<p class=\"whitespace-normal break-words\">First, we need to be assured that such values exist.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>extreme value theorem<\/h3>\n<p>A continuous function [latex]f(x, y)[\/latex] on a closed and bounded set [latex]D[\/latex] in the plane attains an absolute maximum value at some point of [latex]D[\/latex] and an absolute minimum value at some point of [latex]D[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">Now that we know continuous functions on closed, bounded sets attain their extreme values, we need to know how to find them.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>finding absolute extrema<\/h3>\n<p class=\"whitespace-normal break-words\">Assume [latex]z=f(x, y)[\/latex] is a differentiable function of two variables defined on a closed, bounded set [latex]D[\/latex]. Then [latex]f[\/latex] will attain the absolute maximum value and the absolute minimum value, which are the largest and smallest values found among:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The values of [latex]f[\/latex] at the critical points of [latex]f[\/latex] in [latex]D[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">The values of [latex]f[\/latex] on the boundary of [latex]D[\/latex]<\/li>\n<\/ul>\n<ol id=\"fs-id1167793417260\" type=\"i\"><\/ol>\n<\/section>\n<p>The proof follows directly from the extreme value theorem and Fermat&#8217;s theorem. If either extremum is not on the boundary of [latex]D[\/latex], then it&#8217;s at an interior point of [latex]D[\/latex]. But an interior point [latex](x_0, y_0)[\/latex] that&#8217;s an absolute extremum is also a local extremum, so [latex](x_0, y_0)[\/latex] is a critical point by Fermat&#8217;s theorem. Therefore, the only possible values for global extrema are the extreme values on the interior or boundary of [latex]D[\/latex].<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"whitespace-normal break-words\"><strong>Finding Absolute Extrema on a Closed, Bounded Set<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Let [latex]z=f(x, y)[\/latex] be a continuous function defined on a closed, bounded set [latex]D[\/latex], and assume [latex]f[\/latex] is differentiable on [latex]D[\/latex]. To find the absolute maximum and minimum values:<\/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>Find the critical points<\/strong> of [latex]f[\/latex] in [latex]D[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Calculate<\/strong> [latex]f[\/latex] at each critical point<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Determine the maximum and minimum values<\/strong> of [latex]f[\/latex] on the boundary of [latex]D[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Compare all values<\/strong> from steps 2 and 3\u2014the maximum and minimum will occur at one of these values<\/li>\n<\/ol>\n<\/section>\n<p>Finding extrema on the boundary can be challenging. If the boundary is a rectangle or set of straight lines, parameterize the line segments and determine the maxima on each segment. The same approach works for circles and ellipses.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">If the boundary is a more complicated curve defined by [latex]g(x, y)=c[\/latex], and the first-order partial derivatives of [latex]g[\/latex] exist, the method of Lagrange multipliers can help determine extrema on the boundary.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:<\/p>\n<div id=\"fs-id1167793637905\" data-type=\"problem\">\n<div class=\"os-problem-container\">\n<ol id=\"fs-id1167793637915\" type=\"a\">\n<li>[latex]f(x, y)=x^{2}-2xy+4y^{2}-4x-2y+24[\/latex] on the domain defined by\u00a0[latex]0\\leq x\\leq 4[\/latex] and\u00a0[latex]0\\leq y\\leq 2[\/latex].<\/li>\n<li>[latex]g(x, y)=x^{2}+y^{2}+4x-6y[\/latex] on the domain defined by\u00a0[latex]x^{2}+y^{2}\\leq 16[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q147756901\">Show Solution<\/button><\/p>\n<div id=\"q147756901\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. Using the problem-solving strategy, step 1 involves finding the critical points of [latex]f[\/latex] on its domain. Therefore, we first calculate [latex]f_x(x, y)[\/latex] and [latex]f_y(x, y)[\/latex], then set them each equal to zero:<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    f_x(x, y)&=2x-2y-4 \\\\    f_y(x, y)&=-2x+8y-2.    \\end{align}[\/latex]<\/p>\n<p>Setting them equal to zero yields the system of equations<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    2x-2y-4&=0 \\\\    -2x+8y-2&=0.    \\end{align}[\/latex]<\/p>\n<p>The solution to this system is [latex]x=3[\/latex] and [latex]y=1[\/latex]. Therefore [latex](3, 1)[\/latex] is a critical point of [latex]f[\/latex]. Calculating [latex]f(3, 1)[\/latex] gives [latex]f(3, 1)=17[\/latex].\u00a0<span data-type=\"newline\"><br \/>\n<\/span>The next step involves finding the extrema of [latex]f[\/latex] on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:<\/p>\n<figure id=\"attachment_1286\" aria-describedby=\"caption-attachment-1286\" style=\"width: 267px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1286\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135426\/4-7-7.jpeg\" alt=\"A rectangle is drawn in the first quadrant with one corner at the origin, horizontal length 4, and height 2. This rectangle is marked D, and the sides are marked in counterclockwise order from the side overlapping the x axis L1, L2, L3, and L4.\" width=\"267\" height=\"272\" \/><figcaption id=\"caption-attachment-1286\" class=\"wp-caption-text\">Figure 1.\u00a0Graph of the domain of the function [latex]\\small{f(x,y)=x^{2}-2xy+4y^{2}-4x-2y+24}[\/latex].<\/figcaption><\/figure>\n<div class=\"os-problem-container\">\n<p>&nbsp;<\/p>\n<p>[latex]L_1[\/latex] is the line segment connecting\u00a0[latex](0, 0)[\/latex] and\u00a0[latex](4, 0)[\/latex], and it can be parameterized by the equations\u00a0[latex]x(t)=t[\/latex],\u00a0[latex]y(t)=0[\/latex] for\u00a0[latex]0\\leq t\\leq 4[\/latex]. Define\u00a0[latex]g(t)=f(x(t), y(t))[\/latex]. This gives\u00a0[latex]g(t)=t^{2}-4t+24[\/latex]. Differentiating\u00a0[latex]g[\/latex] leads to\u00a0[latex]g'(t)=2t-4[\/latex]. Therefore,\u00a0[latex]g[\/latex] has a critical value at\u00a0[latex]t=2[\/latex], which corresponds to the point\u00a0[latex](2, 0)[\/latex]. Calculating\u00a0[latex]f(2, 0)[\/latex] gives the [latex]z[\/latex]-value\u00a0[latex]20[\/latex].<\/p>\n<p>[latex]L_2[\/latex] is the line segment connecting\u00a0[latex](4, 0)[\/latex] and\u00a0[latex](4, 2)[\/latex], and it can be parameterized by the equations\u00a0[latex]x(t)=4[\/latex],\u00a0[latex]y(t)=t[\/latex] for\u00a0[latex]0\\leq t\\leq 2[\/latex]. Again, define\u00a0[latex]g(t)=f(x(t), y(t))[\/latex]. This gives\u00a0[latex]g(t)=4t^{2}-10t+24[\/latex]. Then,\u00a0[latex]g'(t)=8t-10[\/latex].\u00a0[latex]g[\/latex] has a critical value at\u00a0[latex]t=\\frac{5}{4}[\/latex], which corresponds to the point\u00a0[latex](4, \\frac{5}{4})[\/latex]. Calculating\u00a0[latex]f(4, \\frac{5}{4})[\/latex] gives the [latex]z[\/latex]-value\u00a0[latex]17.75[\/latex].<\/p>\n<p>[latex]L_3[\/latex] is the line segment connecting\u00a0[latex](0, 2)[\/latex] and\u00a0[latex](4, 2)[\/latex], and it can be parameterized by the equations\u00a0[latex]x(t)=t[\/latex],\u00a0[latex]y(t)=2[\/latex] for\u00a0[latex]0\\leq t\\leq 4[\/latex]. Again, define\u00a0[latex]g(t)=f(x(t), y(t))[\/latex]. This gives\u00a0[latex]g(t)=t^{2}-8t+36[\/latex]. The critical value corresponds to the point\u00a0[latex](4, 2)[\/latex]. So, calculating\u00a0[latex]f(4, 2)[\/latex] gives the [latex]z[\/latex]-value\u00a0[latex]20[\/latex].<\/p>\n<p>[latex]L_4[\/latex] is the line segment connecting\u00a0[latex](0, 0)[\/latex] and\u00a0[latex](0, 2)[\/latex], and it can be parameterized by the equations\u00a0[latex]x(t)=0[\/latex],\u00a0[latex]y(t)=t[\/latex] for\u00a0[latex]0\\leq t\\leq 2[\/latex]. This time [latex]g(t)=4t^{2}-2t+24[\/latex] and the critical value\u00a0[latex]t=\\frac{1}{4}[\/latex]corresponds to the point\u00a0[latex](0, \\frac{1}{4})[\/latex]. Calculating\u00a0[latex]f(0, \\frac{1}{4})[\/latex] gives the [latex]z[\/latex]-value\u00a0[latex]23.75[\/latex]. We also need to find the values of [latex]f(x, y)[\/latex] at the corners of its domain. These corners are located at [latex](0, 0)[\/latex],\u00a0[latex](4, 0)[\/latex],\u00a0[latex](4, 2)[\/latex], and\u00a0[latex](0, 2)[\/latex]:<br \/>\n[latex]\\hspace{5cm}\\begin{align}  f(0,0)&=(0)^2-2(0)(0)+4(0)^2-4(0)-2(0)+24=24 \\\\  f(4,0)&=(4)^2-2(4)(0)+4(0)^2-4(4)-2(0)+24=24 \\\\  f(4,2)&=(4)^2-2(4)(2)+4(2)^2-4(4)-2(2)+24=20 \\\\  f(0,2)&=(0)^2-2(0)(2)+4(2)^2-4(0)-2(2)+24=36.  \\end{align}[\/latex]<\/p>\n<p>The absolute maximum value is [latex]36[\/latex], which occurs at [latex](0, 2)[\/latex], and the global minimum value is [latex]17[\/latex], which occurs at [latex](3, 1)[\/latex] as shown in the following figure.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_1287\" aria-describedby=\"caption-attachment-1287\" style=\"width: 729px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1287\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135537\/4-7-8.jpeg\" alt=\"The function f(x, y) = x2 \u2212 2xy \u2013 4x + 4y2 \u2013 2y + 24 is shown with local minima at (4, 2, 20) and (2, 0, 20) and local maximum at (0, 2, 36). The shape is a plane curving up on the corners significantly near (0, 2) and slightly less near (4, 0).\" width=\"729\" height=\"587\" \/><figcaption id=\"caption-attachment-1287\" class=\"wp-caption-text\">Figure 2. The function [latex]\\small{f(x,y)}[\/latex] has one global minimum and one global maximum over its domain.<\/figcaption><\/figure>\n<p>b.\u00a0Using the problem-solving strategy, step 1 involves finding the critical points of [latex]g[\/latex] on its domain. Therefore, we first calculate [latex]g_x(x, y)[\/latex] and [latex]g_y(x, y)[\/latex], then set them each equal to zero:<\/p><\/div>\n<div><\/div>\n<div>\n<p>[latex]\\hspace{8cm}\\begin{align}g_x(x, y)&=2y+4 \\\\g_y(x, y)&=2y-6.    \\end{align}[\/latex]<\/p>\n<p>Setting them equal to zero yields the system of equations<\/p>\n<\/div>\n<p>[latex]\\hspace{8cm}\\begin{align}    2x+4&=0 \\\\  2y-6&=0.  \\end{align}[\/latex]<br \/>\nThe solution to this system is [latex]x=-2[\/latex] and [latex]y=3[\/latex]. Therefore, [latex](-2, 3)[\/latex] is a critical point of [latex]g[\/latex]. Calculating [latex]g(-2, 3)[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]g(-2,3)=(-2)^2+3^2+4(-2)-6(3)=4+9-8-18=-13.[\/latex]<\/p>\n<p>The next step involves finding the extrema of [latex]g[\/latex]\u00a0on the boundary of its domain. The boundary of its domain consists of a circle of radius [latex]4[\/latex] centered at the origin as shown in the following graph.<\/p>\n<figure id=\"attachment_1288\" aria-describedby=\"caption-attachment-1288\" style=\"width: 417px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1288\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135649\/4-7-9.jpeg\" alt=\"A filled-in circle marked D of radius four with center at the origin.\" width=\"417\" height=\"423\" \/><figcaption id=\"caption-attachment-1288\" class=\"wp-caption-text\">Figure 3.\u00a0Graph of the domain of the function [latex]\\small{g(x,y)=x^{2}+y^{2}+4x-6y}[\/latex].<\/figcaption><\/figure>\n<p>The boundary of the domain of [latex]g[\/latex] can be parameterized using the functions\u00a0[latex]x(t)=4\\cos t[\/latex],\u00a0[latex]y(t)=4\\sin t[\/latex] for\u00a0[latex]0\\leq{t}\\leq2\\pi[\/latex]. Define\u00a0[latex]h(t)=g(x(t), y(t))[\/latex]:<br \/>\n[latex]\\hspace{5cm}\\begin{align}  h(t)&=g(x(t),y(t)) \\\\  &=(4\\cos{t})^2+(4\\sin t)^2+4(4\\cos{t})-6(4\\sin t) \\\\  &=16\\cos^2{t}+16\\sin^2{t}+16\\cos{t}-24\\sin t \\\\  &=16+16\\cos{t}-24\\sin t.\\end{align}[\/latex]<br \/>\nSetting\u00a0[latex]h'(t)=0[\/latex] leads to<br \/>\n[latex]\\hspace{5cm}\\begin{align}-16\\sin{t}-24\\cos{t}&=0 \\\\-16\\sin{t}&=24\\cos{t} \\\\\\frac{-16\\sin{t}}{-16\\cos{t}}&=\\frac{24\\cos{t}}{-16\\cos{t}} \\\\\\tan{t}&=-\\frac32\\end{align}[\/latex]<br \/>\nThis equation has two solutions over the interval\u00a0[latex]0\\leq{t}\\leq2\\pi[\/latex]. One is\u00a0[latex]t=\\pi-\\arctan{(\\frac32)}[\/latex] and the other is\u00a0[latex]t=2\\pi-\\arctan{(\\frac32)}[\/latex]. For the first angle,<br \/>\n[latex]\\hspace{4cm}\\begin{align}\\sin{t}&=\\sin{(\\pi-\\arctan{(\\frac32)})}=\\sin{(\\arctan{(\\frac32)})}=\\frac{3\\sqrt{13}}{13} \\\\\\cos{t}&=\\cos{(\\pi-\\arctan{(\\frac32)})}=-\\cos{(\\arctan{(\\frac32)})}=-\\frac{2\\sqrt{13}}{13}\\end{align}[\/latex]<br \/>\nTherefore, [latex]x(t)=4\\cos{t}=\\frac{8\\sqrt{13}}{13}[\/latex] and [latex]y(t)=4\\sin{t}=-\\frac{12\\sqrt{13}}{13}[\/latex], so [latex]\\left(\\frac{8\\sqrt{13}}{13},-\\frac{12\\sqrt{13}}{13}\\right)[\/latex] is a critical point on the boundary and[latex]\\hspace{4cm}\\begin{align}g\\left(-\\frac{8\\sqrt{13}}{13},\\frac{12\\sqrt{13}}{13}\\right)&=\\left(-\\frac{8\\sqrt{13}}{13}\\right)^2+\\left(\\frac{12\\sqrt{13}}{13}\\right)^2+4\\left(-\\frac{8\\sqrt{13}}{13}\\right)-6\\left(\\frac{12\\sqrt{13}}{13}\\right) \\\\&=\\frac{144}{13}+\\frac{64}{13}-\\frac{32\\sqrt{13}}{13}-\\frac{72\\sqrt{13}}{13} \\\\&=\\frac{208-104\\sqrt{13}}{13}\\approx-12.844.\\end{align}[\/latex]<br \/>\nFor the second angle,<br \/>\n[latex]\\hspace{5cm}\\begin{align}\\sin{t}&=\\sin{(2\\pi-\\arctan{(\\frac32)})}=-\\sin{(\\arctan{(\\frac32)})}=-\\frac{3\\sqrt{13}}{13} \\\\\\cos{t}&=\\cos{(2\\pi-\\arctan{(\\frac32)})}=\\cos{(\\arctan{(\\frac32)})}=\\frac{2\\sqrt{13}}{13}.\\end{align}[\/latex]<br \/>\nTherefore,\u00a0[latex]x(t)=4\\cos{t}=\\frac{8\\sqrt{13}}{13}[\/latex] and\u00a0[latex]y(t)=4\\sin{t}=-\\frac{12\\sqrt{13}}{13}[\/latex], so\u00a0[latex]\\left(\\frac{8\\sqrt{13}}{13},-\\frac{12\\sqrt{13}}{13}\\right)[\/latex] is a critical point on the boundary and[latex]\\hspace{4cm}\\begin{align}g\\left(\\frac{8\\sqrt{13}}{13},-\\frac{12\\sqrt{13}}{13}\\right)&=\\left(\\frac{8\\sqrt{13}}{13}\\right)^2+\\left(-\\frac{12\\sqrt{13}}{13}\\right)^2+4\\left(\\frac{8\\sqrt{13}}{13}\\right)-6\\left(-\\frac{12\\sqrt{13}}{13}\\right) \\\\&=\\frac{144}{13}+\\frac{64}{13}+\\frac{32\\sqrt{13}}{13}+\\frac{72\\sqrt{13}}{13} \\\\&=\\frac{208+104\\sqrt{13}}{13}\\approx44.844.\\end{align}[\/latex]The absolute minimum of [latex]g[\/latex] is [latex]-13[\/latex] which is attained at the point [latex](-2, 3)[\/latex], which is an interior point of [latex]D[\/latex]. The absolute maximum of [latex]g[\/latex] is approximately equal to [latex]44.844[\/latex], which is attained at the boundary point [latex]\\left(\\frac{8\\sqrt{13}}{13},-\\frac{12\\sqrt{13}}{13}\\right)[\/latex]. These are the absolute extrema of [latex]g[\/latex] on [latex]D[\/latex] as shown in the following figure.<\/p>\n<figure id=\"attachment_1289\" aria-describedby=\"caption-attachment-1289\" style=\"width: 688px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1289\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135817\/4-7-10.jpeg\" alt=\"The function f(x, y) = x2 + 4x + y2 \u2013 6y is shown with local minimum at (\u201312\/5, 16\/5, \u201364\/5) and local maximum at (12\/5, \u221216\/5, 224\/5). The shape is a plane curving up from near (\u22124, 4) to (4, \u22124).\" width=\"688\" height=\"470\" \/><figcaption id=\"caption-attachment-1289\" class=\"wp-caption-text\">Figure 4. The function [latex]\\small{f(x,y)}[\/latex] has a local minimum and a local maximum.<\/figcaption><\/figure>\n<\/div>\n<div class=\"os-problem-container\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":6,"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-1167","chapter","type-chapter","status-publish","hentry"],"part":430,"module-header":"- Select Header -","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\/1167","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":4,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1167\/revisions"}],"predecessor-version":[{"id":1829,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1167\/revisions\/1829"}],"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\/1167\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1167"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1167"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1167"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1167"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}