{"id":1166,"date":"2025-10-15T14:54:01","date_gmt":"2025-10-15T14:54:01","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1166"},"modified":"2025-10-29T12:17:34","modified_gmt":"2025-10-29T12:17:34","slug":"critical-points-and-local-extrema-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/critical-points-and-local-extrema-learn-it-2\/","title":{"raw":"Critical Points and Local Extrema: Learn It 2","rendered":"Critical Points and Local Extrema: Learn It 2"},"content":{"raw":"<h2>Second Derivative Test<\/h2>\r\n<p class=\"whitespace-normal break-words\">Consider the function [latex]f(x)=x^{3}[\/latex]. This function has a critical point at [latex]x=0[\/latex] since [latex]f'(0)=3(0)^{2}=0[\/latex]. However, [latex]f[\/latex] does not have an extreme value at [latex]x=0[\/latex]. The existence of a critical point at [latex]x=x_0[\/latex] does not guarantee a local extremum there.<\/p>\r\n<p class=\"whitespace-normal break-words\">The same is true for functions of two or more variables. One way this can happen is at a saddle point. Here's an example:<\/p>\r\n\r\n[caption id=\"attachment_1279\" align=\"aligncenter\" width=\"483\"]<img class=\"size-full wp-image-1279\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134416\/4-7-3.jpeg\" alt=\"x^{2}-y^{2}\" width=\"483\" height=\"651\" \/> Figure 1. Graph of the function\u00a0[latex]\\small{z=x^{2}-y^{2}}[\/latex].\u00a0This graph has a saddle point at the origin.[\/caption]In this graph, the origin is a saddle point. The first partial derivatives of [latex]f(x, y)=x^{2}-y^{2}[\/latex] are both zero at this point, but it's neither a maximum nor a minimum for the function. The vertical trace corresponding to [latex]y=0[\/latex] is [latex]z=x^{2}[\/latex] (a parabola opening upward), but the vertical trace corresponding to [latex]x=0[\/latex] is [latex]z=-y^{2}[\/latex] (a parabola opening downward). Therefore, it's both a global maximum for one trace and a global minimum for another.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>saddle point<\/h3>\r\nGiven the function [latex]z=f(x, y)[\/latex], the point [latex](x_0, y_0, f(x_0, y_0))[\/latex] is a <strong>saddle point<\/strong> if both [latex]f_x(x_0, y_0)=0[\/latex] and [latex]f_y(x_0, y_0)=0[\/latex], but [latex]f[\/latex] does not have a local extremum at [latex](x_0, y_0)[\/latex].\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">The second derivative test for single-variable functions provides a method for determining whether an extremum occurs at a critical point. Let's recall how it works before extending it to two variables.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Second Derivative Test (Single Variable Version)<\/strong>Suppose that [latex] f(x) [\/latex] is twice-differentiable at [latex] x = x_0 [\/latex] and [latex] f'(x_0) = 0 [\/latex].\r\n<ul>\r\n \t<li>If [latex] f''(x_0) &lt; 0 [\/latex], then [latex]f [\/latex] has a local maximum at [latex] x_0 [\/latex].<\/li>\r\n \t<li>If [latex] f''(x_0) &gt; 0 [\/latex], then [latex]f [\/latex] has a local minimum at [latex] x_0 [\/latex].<\/li>\r\n \t<li>If [latex] f''(x_0) = 0 [\/latex], then the test is inconclusive.<\/li>\r\n<\/ul>\r\n<\/section>When extending this to functions of two variables, an issue arises: there are four different second-order partial derivatives (though equality of mixed partials reduces this to three). The second derivative test for a function of two variables, stated in the following theorem, uses a\u00a0<strong><span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term198\" data-type=\"term\">discriminant [latex]D[\/latex]\u00a0<\/span><\/strong>that replaces [latex]f''(x_0)[\/latex] in the second derivative test for a function of one variable.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>second derivative test<\/h3>\r\nLet [latex]z=f(x, y)[\/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[\/latex]. Suppose [latex]f_x(x_0, y_0)=0[\/latex] and [latex]f_y(x_0, y_0)=0[\/latex]. Define the quantity\r\n<p style=\"text-align: center;\">[latex]\\large{D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2}.[\/latex]<\/p>\r\n\r\n<ul id=\"fs-id1167793417260\" type=\"i\">\r\n \t<li>If\u00a0[latex]D&gt;0[\/latex] and\u00a0[latex]f_{xx}(x_0,y_0)&gt;0[\/latex], then\u00a0[latex]f[\/latex] has a local minimum at\u00a0[latex](x_0, y_0)[\/latex].<\/li>\r\n \t<li>If\u00a0[latex]D&gt;0[\/latex] and\u00a0[latex]f_{xx}(x_0,y_0)&lt;0[\/latex], then\u00a0[latex]f[\/latex] has a local maximum at\u00a0[latex](x_0,y_0)[\/latex].<\/li>\r\n \t<li>If\u00a0[latex]D&lt;0[\/latex], then\u00a0[latex]f[\/latex] has a saddle point at\u00a0[latex](x_0, y_0)[\/latex].<\/li>\r\n \t<li>If\u00a0[latex]D=0[\/latex], then the test is inconclusive.<\/li>\r\n<\/ul>\r\nSee\u00a0Figure 2.\r\n\r\n<\/section>\r\n\r\n[caption id=\"attachment_1281\" align=\"aligncenter\" width=\"943\"]<img class=\"size-full wp-image-1281\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134549\/4-7-4.jpeg\" alt=\"This figure consists of three figures labeled a, b, and c. Figure a has two bulbous mounds pointing down, and the two extrema are listed as the local minima. Figure b has two bulbous mounds pointed up, and the two extrema are listed as the local maxima. Figure c is shaped like a saddle, and in the middle of the saddle, a point is marked as the saddle point.\" width=\"943\" height=\"345\" \/> Figure 2. The second derivative test can often determine whether a function of two variables has a local minima (a), a local maxima (b), or a saddle point (c).[\/caption]\r\n\r\nTo apply the second derivative test, we first need to find the critical points of the function. Here's a systematic approach.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>Using the Second Derivative Test for Functions of Two Variables<\/strong>\r\n<p class=\"whitespace-normal break-words\">Let [latex]z=f(x,y)[\/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[\/latex]. To apply the second derivative test to find local extrema:<\/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 critical points<\/strong> [latex](x_0, y_0)[\/latex] where [latex]f_x(x_0, y_0)=f_y(x_0, y_0)=0[\/latex]. Discard any points where at least one of the partial derivatives does not exist.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Calculate the discriminant<\/strong> [latex]D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2[\/latex] for each critical point.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Apply the second derivative test<\/strong> to determine whether each critical point is a local maximum, local minimum, saddle point, or whether the test is inconclusive.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793244016\">Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:<\/p>\r\n\r\n<ol id=\"fs-id1167793374827\" type=\"a\">\r\n \t<li>[latex]f(x, y)=4x^{2}+9y^{2}+8x-36y+24[\/latex]<\/li>\r\n \t<li>[latex]g(x, y)=\\frac{1}{3}x^{3}+y^{2}+2xy-6x-3y+4[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"914523760\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"914523760\"]\r\n\r\na. Step 1 of the problem-solving strategy involves finding the critical points of [latex]f[\/latex]. To do this, we first calculate [latex]f_x(x, y)[\/latex] and [latex]f_y(x, y)[\/latex], then set each of them equal to zero:\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\nf_x(x,y)&amp;=8x+8 \\\\\r\n\r\nf_y(x,y)&amp;=18y-36.\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\n8x+8&amp;=0 \\\\\r\n\r\n18y-36&amp;=0.\r\n\r\n\\end{align}[\/latex]\r\n\r\nThe solution to this system is [latex]x=-1[\/latex] and [latex]y=2[\/latex]. Therefore [latex](-1, 2)[\/latex] is a critical point of [latex]f[\/latex].<span data-type=\"newline\">\r\n<\/span>Step 2 of the problem-solving strategy involves calculating [latex]D[\/latex]. To do this, we first calculate the second partial derivatives of [latex]f[\/latex]:<span data-type=\"newline\">\r\n<\/span>\r\n\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\nf_{xx}(x,y)&amp;=8 \\\\\r\n\r\nf_{xy}(x,y)&amp;=0 \\\\\r\n\r\nf_{yy}(x,y)&amp;=18.\r\n\r\n\\end{align}[\/latex]\r\n\r\nTherefore, [latex]D=f_{xx}(-1,2)f_{yy}(-1,2)-(f_{xy}(-1,2))^2=(8)(18)-(0)^2=144[\/latex].\r\n\r\nStep 3 states to check the\u00a0Second Derivative Test for Functions of Two Variables.. Since [latex]D&gt;0[\/latex] and [latex]f_{xx}(-1,2)&gt;0[\/latex], this corresponds to case 1. Therefore, [latex]f[\/latex] has a local minimum at [latex](-1, 2)[\/latex] as shown in the following figure.\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1283\" align=\"aligncenter\" width=\"813\"]<img class=\"wp-image-1283 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135113\/4-7-5.jpeg\" alt=\"The function f(x, y) = 4x2 + 9y2 + 8x \u2013 36y + 24 is shown with local minimum at (\u20131, 2, \u201316). The shape is a plane curving up on both ends parallel to the y axis.\" width=\"813\" height=\"405\" \/> Figure 3. The function [latex]\\small{f(x,y)}[\/latex] has a local minimum at [latex]\\small{(-1,2,-16)}[\/latex].[\/caption]b. For step 1, we first calculate [latex]g_x(x,y)[\/latex] and [latex]g_y(x,y)[\/latex], then set each of them equal to zero:[latex]\\hspace{8cm}\\begin{align}g_x(x,y)&amp;=x^2+2y-6 \\\\g_y(x,y)&amp;=2y+2x-3.\\end{align}[\/latex]\r\nSetting them equal to zero yields the system of equations\r\n[latex]\\hspace{8cm}\\begin{align}x^2+2y-6&amp;=0 \\\\2y+2x-3&amp;=0.\\end{align}[\/latex]\r\nTo solve this system, first solve the second equation for [latex]y[\/latex]. This gives [latex]y=\\frac{3-2x}{2}[\/latex]. Substituting this into the first equation gives[latex]\\hspace{7.5cm}\\begin{align}x^2+3-2x-6&amp;=0 \\\\x^2-2x-3&amp;=0 \\\\(x-3)(x+1)&amp;=0.\\end{align}[\/latex]\r\nTherefore, [latex]x=-1[\/latex] or [latex]x=3[\/latex]. Substituting these values into the equation\u00a0[latex]y=\\frac{3-2x}{2}[\/latex] yields the critical points [latex](-1,\\frac52)[\/latex] and [latex](3,-\\frac32)[\/latex].\r\n<p id=\"fs-id1167794138811\">Step 2 involves calculating the second partial derivatives of [latex]g[\/latex]:<\/p>\r\n[latex]\\hspace{8cm}\\begin{align}\r\n\r\ng_{xx}(x,y)&amp;=2x \\\\\r\n\r\ng_{xy}(x,y)&amp;=2 \\\\\r\n\r\ng_{yy}(x,y)&amp;=2.\r\n\r\n\\end{align}[\/latex]\r\n\r\nThen, we find a general formula for\u00a0[latex]D[\/latex]:\r\n\r\n[latex]\\hspace{7cm}\\begin{align}\r\n\r\nD&amp;=g_{xx}(x_0,y_0)g_{yy}(x_0,y_0)-(g_{xy}(x_0,y_0))^2 \\\\\r\n\r\n&amp;=(2x_0)(2)-2^2 \\\\\r\n\r\n&amp;=4x_0-4.\r\n\r\n\\end{align}[\/latex]\r\n\r\nNext, we substitute each critical point into this formula:\r\n\r\n[latex]\\hspace{7cm}\\begin{align}\r\n\r\nD(-1,\\frac52)&amp;=(2(-1))(2)-(2)^2=-4-4=-8 \\\\\r\n\r\nD(3,-\\frac32)&amp;=(2(3))(2)-(2)^2=12-4=8.\r\n\r\n\\end{align}[\/latex]\r\n\r\nIn step 3, we note that, applying the Second Derivative Test for Functions of Two Variables to point [latex](-1,\\frac{5}{2})[\/latex] leads to case 3, which means that [latex](-1,\\frac{5}{2})[\/latex]\u00a0is a saddle point. Applying the theorem to point [latex](3,-\\frac{3}{2})[\/latex]\u00a0leads to case 1, which means that [latex](3,-\\frac{3}{2})[\/latex] corresponds to a local minimum as shown in the following figure.\r\n\r\n[caption id=\"attachment_1284\" align=\"aligncenter\" width=\"829\"]<img class=\"size-full wp-image-1284\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135248\/4-7-6.jpeg\" alt=\"The function f(x, y) = (1\/3)x3 + y2 + + 2xy \u2013 6x \u2013 3y + 4 is shown with local minimum at (3, \u20133\/2, \u201329\/4) and saddle point at (\u22121, 5\/2, 41\/12). The shape is a plane curving up on the corners near (4, 3) and (\u22122, \u22122).\" width=\"829\" height=\"488\" \/> Figure 4. The function [latex]\\small{g(x,y)}[\/latex] has a local minimum and a saddle point.[\/caption][\/hidden-answer]<\/section>","rendered":"<h2>Second Derivative Test<\/h2>\n<p class=\"whitespace-normal break-words\">Consider the function [latex]f(x)=x^{3}[\/latex]. This function has a critical point at [latex]x=0[\/latex] since [latex]f'(0)=3(0)^{2}=0[\/latex]. However, [latex]f[\/latex] does not have an extreme value at [latex]x=0[\/latex]. The existence of a critical point at [latex]x=x_0[\/latex] does not guarantee a local extremum there.<\/p>\n<p class=\"whitespace-normal break-words\">The same is true for functions of two or more variables. One way this can happen is at a saddle point. Here&#8217;s an example:<\/p>\n<figure id=\"attachment_1279\" aria-describedby=\"caption-attachment-1279\" style=\"width: 483px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1279\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134416\/4-7-3.jpeg\" alt=\"x^{2}-y^{2}\" width=\"483\" height=\"651\" \/><figcaption id=\"caption-attachment-1279\" class=\"wp-caption-text\">Figure 1. Graph of the function\u00a0[latex]\\small{z=x^{2}-y^{2}}[\/latex].\u00a0This graph has a saddle point at the origin.<\/figcaption><\/figure>\n<p>In this graph, the origin is a saddle point. The first partial derivatives of [latex]f(x, y)=x^{2}-y^{2}[\/latex] are both zero at this point, but it&#8217;s neither a maximum nor a minimum for the function. The vertical trace corresponding to [latex]y=0[\/latex] is [latex]z=x^{2}[\/latex] (a parabola opening upward), but the vertical trace corresponding to [latex]x=0[\/latex] is [latex]z=-y^{2}[\/latex] (a parabola opening downward). Therefore, it&#8217;s both a global maximum for one trace and a global minimum for another.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>saddle point<\/h3>\n<p>Given the function [latex]z=f(x, y)[\/latex], the point [latex](x_0, y_0, f(x_0, y_0))[\/latex] is a <strong>saddle point<\/strong> if both [latex]f_x(x_0, y_0)=0[\/latex] and [latex]f_y(x_0, y_0)=0[\/latex], but [latex]f[\/latex] does not have a local extremum at [latex](x_0, y_0)[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">The second derivative test for single-variable functions provides a method for determining whether an extremum occurs at a critical point. Let&#8217;s recall how it works before extending it to two variables.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Second Derivative Test (Single Variable Version)<\/strong>Suppose that [latex]f(x)[\/latex] is twice-differentiable at [latex]x = x_0[\/latex] and [latex]f'(x_0) = 0[\/latex].<\/p>\n<ul>\n<li>If [latex]f''(x_0) < 0[\/latex], then [latex]f[\/latex] has a local maximum at [latex]x_0[\/latex].<\/li>\n<li>If [latex]f''(x_0) > 0[\/latex], then [latex]f[\/latex] has a local minimum at [latex]x_0[\/latex].<\/li>\n<li>If [latex]f''(x_0) = 0[\/latex], then the test is inconclusive.<\/li>\n<\/ul>\n<\/section>\n<p>When extending this to functions of two variables, an issue arises: there are four different second-order partial derivatives (though equality of mixed partials reduces this to three). The second derivative test for a function of two variables, stated in the following theorem, uses a\u00a0<strong><span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term198\" data-type=\"term\">discriminant [latex]D[\/latex]\u00a0<\/span><\/strong>that replaces [latex]f''(x_0)[\/latex] in the second derivative test for a function of one variable.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>second derivative test<\/h3>\n<p>Let [latex]z=f(x, y)[\/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[\/latex]. Suppose [latex]f_x(x_0, y_0)=0[\/latex] and [latex]f_y(x_0, y_0)=0[\/latex]. Define the quantity<\/p>\n<p style=\"text-align: center;\">[latex]\\large{D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2}.[\/latex]<\/p>\n<ul id=\"fs-id1167793417260\" type=\"i\">\n<li>If\u00a0[latex]D>0[\/latex] and\u00a0[latex]f_{xx}(x_0,y_0)>0[\/latex], then\u00a0[latex]f[\/latex] has a local minimum at\u00a0[latex](x_0, y_0)[\/latex].<\/li>\n<li>If\u00a0[latex]D>0[\/latex] and\u00a0[latex]f_{xx}(x_0,y_0)<0[\/latex], then\u00a0[latex]f[\/latex] has a local maximum at\u00a0[latex](x_0,y_0)[\/latex].<\/li>\n<li>If\u00a0[latex]D<0[\/latex], then\u00a0[latex]f[\/latex] has a saddle point at\u00a0[latex](x_0, y_0)[\/latex].<\/li>\n<li>If\u00a0[latex]D=0[\/latex], then the test is inconclusive.<\/li>\n<\/ul>\n<p>See\u00a0Figure 2.<\/p>\n<\/section>\n<figure id=\"attachment_1281\" aria-describedby=\"caption-attachment-1281\" style=\"width: 943px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1281\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134549\/4-7-4.jpeg\" alt=\"This figure consists of three figures labeled a, b, and c. Figure a has two bulbous mounds pointing down, and the two extrema are listed as the local minima. Figure b has two bulbous mounds pointed up, and the two extrema are listed as the local maxima. Figure c is shaped like a saddle, and in the middle of the saddle, a point is marked as the saddle point.\" width=\"943\" height=\"345\" \/><figcaption id=\"caption-attachment-1281\" class=\"wp-caption-text\">Figure 2. The second derivative test can often determine whether a function of two variables has a local minima (a), a local maxima (b), or a saddle point (c).<\/figcaption><\/figure>\n<p>To apply the second derivative test, we first need to find the critical points of the function. Here&#8217;s a systematic approach.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>Using the Second Derivative Test for Functions of Two Variables<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Let [latex]z=f(x,y)[\/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[\/latex]. To apply the second derivative test to find local extrema:<\/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 critical points<\/strong> [latex](x_0, y_0)[\/latex] where [latex]f_x(x_0, y_0)=f_y(x_0, y_0)=0[\/latex]. Discard any points where at least one of the partial derivatives does not exist.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Calculate the discriminant<\/strong> [latex]D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2[\/latex] for each critical point.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Apply the second derivative test<\/strong> to determine whether each critical point is a local maximum, local minimum, saddle point, or whether the test is inconclusive.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793244016\">Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:<\/p>\n<ol id=\"fs-id1167793374827\" type=\"a\">\n<li>[latex]f(x, y)=4x^{2}+9y^{2}+8x-36y+24[\/latex]<\/li>\n<li>[latex]g(x, y)=\\frac{1}{3}x^{3}+y^{2}+2xy-6x-3y+4[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q914523760\">Show Solution<\/button><\/p>\n<div id=\"q914523760\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. Step 1 of the problem-solving strategy involves finding the critical points of [latex]f[\/latex]. To do this, we first calculate [latex]f_x(x, y)[\/latex] and [latex]f_y(x, y)[\/latex], then set each of them equal to zero:<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    f_x(x,y)&=8x+8 \\\\    f_y(x,y)&=18y-36.    \\end{align}[\/latex]<\/p>\n<p>Setting them equal to zero yields the system of equations<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    8x+8&=0 \\\\    18y-36&=0.    \\end{align}[\/latex]<\/p>\n<p>The solution to this system is [latex]x=-1[\/latex] and [latex]y=2[\/latex]. Therefore [latex](-1, 2)[\/latex] is a critical point of [latex]f[\/latex].<span data-type=\"newline\"><br \/>\n<\/span>Step 2 of the problem-solving strategy involves calculating [latex]D[\/latex]. To do this, we first calculate the second partial derivatives of [latex]f[\/latex]:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    f_{xx}(x,y)&=8 \\\\    f_{xy}(x,y)&=0 \\\\    f_{yy}(x,y)&=18.    \\end{align}[\/latex]<\/p>\n<p>Therefore, [latex]D=f_{xx}(-1,2)f_{yy}(-1,2)-(f_{xy}(-1,2))^2=(8)(18)-(0)^2=144[\/latex].<\/p>\n<p>Step 3 states to check the\u00a0Second Derivative Test for Functions of Two Variables.. Since [latex]D>0[\/latex] and [latex]f_{xx}(-1,2)>0[\/latex], this corresponds to case 1. Therefore, [latex]f[\/latex] has a local minimum at [latex](-1, 2)[\/latex] as shown in the following figure.<\/p>\n<p>&nbsp;<\/p>\n<figure id=\"attachment_1283\" aria-describedby=\"caption-attachment-1283\" style=\"width: 813px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1283 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135113\/4-7-5.jpeg\" alt=\"The function f(x, y) = 4x2 + 9y2 + 8x \u2013 36y + 24 is shown with local minimum at (\u20131, 2, \u201316). The shape is a plane curving up on both ends parallel to the y axis.\" width=\"813\" height=\"405\" \/><figcaption id=\"caption-attachment-1283\" class=\"wp-caption-text\">Figure 3. The function [latex]\\small{f(x,y)}[\/latex] has a local minimum at [latex]\\small{(-1,2,-16)}[\/latex].<\/figcaption><\/figure>\n<p>b. For step 1, we first calculate [latex]g_x(x,y)[\/latex] and [latex]g_y(x,y)[\/latex], then set each of them equal to zero:[latex]\\hspace{8cm}\\begin{align}g_x(x,y)&=x^2+2y-6 \\\\g_y(x,y)&=2y+2x-3.\\end{align}[\/latex]<br \/>\nSetting them equal to zero yields the system of equations<br \/>\n[latex]\\hspace{8cm}\\begin{align}x^2+2y-6&=0 \\\\2y+2x-3&=0.\\end{align}[\/latex]<br \/>\nTo solve this system, first solve the second equation for [latex]y[\/latex]. This gives [latex]y=\\frac{3-2x}{2}[\/latex]. Substituting this into the first equation gives[latex]\\hspace{7.5cm}\\begin{align}x^2+3-2x-6&=0 \\\\x^2-2x-3&=0 \\\\(x-3)(x+1)&=0.\\end{align}[\/latex]<br \/>\nTherefore, [latex]x=-1[\/latex] or [latex]x=3[\/latex]. Substituting these values into the equation\u00a0[latex]y=\\frac{3-2x}{2}[\/latex] yields the critical points [latex](-1,\\frac52)[\/latex] and [latex](3,-\\frac32)[\/latex].<\/p>\n<p id=\"fs-id1167794138811\">Step 2 involves calculating the second partial derivatives of [latex]g[\/latex]:<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}    g_{xx}(x,y)&=2x \\\\    g_{xy}(x,y)&=2 \\\\    g_{yy}(x,y)&=2.    \\end{align}[\/latex]<\/p>\n<p>Then, we find a general formula for\u00a0[latex]D[\/latex]:<\/p>\n<p>[latex]\\hspace{7cm}\\begin{align}    D&=g_{xx}(x_0,y_0)g_{yy}(x_0,y_0)-(g_{xy}(x_0,y_0))^2 \\\\    &=(2x_0)(2)-2^2 \\\\    &=4x_0-4.    \\end{align}[\/latex]<\/p>\n<p>Next, we substitute each critical point into this formula:<\/p>\n<p>[latex]\\hspace{7cm}\\begin{align}    D(-1,\\frac52)&=(2(-1))(2)-(2)^2=-4-4=-8 \\\\    D(3,-\\frac32)&=(2(3))(2)-(2)^2=12-4=8.    \\end{align}[\/latex]<\/p>\n<p>In step 3, we note that, applying the Second Derivative Test for Functions of Two Variables to point [latex](-1,\\frac{5}{2})[\/latex] leads to case 3, which means that [latex](-1,\\frac{5}{2})[\/latex]\u00a0is a saddle point. Applying the theorem to point [latex](3,-\\frac{3}{2})[\/latex]\u00a0leads to case 1, which means that [latex](3,-\\frac{3}{2})[\/latex] corresponds to a local minimum as shown in the following figure.<\/p>\n<figure id=\"attachment_1284\" aria-describedby=\"caption-attachment-1284\" style=\"width: 829px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1284\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23135248\/4-7-6.jpeg\" alt=\"The function f(x, y) = (1\/3)x3 + y2 + + 2xy \u2013 6x \u2013 3y + 4 is shown with local minimum at (3, \u20133\/2, \u201329\/4) and saddle point at (\u22121, 5\/2, 41\/12). The shape is a plane curving up on the corners near (4, 3) and (\u22122, \u22122).\" width=\"829\" height=\"488\" \/><figcaption id=\"caption-attachment-1284\" class=\"wp-caption-text\">Figure 4. The function [latex]\\small{g(x,y)}[\/latex] has a local minimum and a saddle point.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":5,"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-1166","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\/1166","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\/1166\/revisions"}],"predecessor-version":[{"id":1828,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1166\/revisions\/1828"}],"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\/1166\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1166"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1166"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1166"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1166"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}