{"id":652,"date":"2025-10-09T15:28:13","date_gmt":"2025-10-09T15:28:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=652"},"modified":"2025-10-29T11:55:58","modified_gmt":"2025-10-29T11:55:58","slug":"critical-points-and-local-extrema-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/critical-points-and-local-extrema-learn-it-1\/","title":{"raw":"Critical Points and Local Extrema: Learn It 1","rendered":"Critical Points and Local Extrema: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find critical points using partial derivatives<\/li>\r\n \t<li>Apply the second derivative test<\/li>\r\n \t<li>Find absolute extrema on closed regions<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Critical Points<\/h2>\r\n<p class=\"whitespace-normal break-words\">One of the most useful applications of derivatives for single-variable functions is finding maximum and minimum values. This application is equally important for functions of two or more variables. The main ideas\u2014finding critical points and using derivative tests\u2014are still valid, but more independent variables lead to more possible outcomes.<\/p>\r\n<p class=\"whitespace-normal break-words\">For functions of a single variable, critical points occur where the derivative equals zero or does not exist. For functions of two or more variables, the concept is the same, except we work with partial derivatives.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>critical point<\/h3>\r\nLet [latex]z=f(x, y)[\/latex] be a function of two variables that is defined on an open set containing the point [latex](x_0, y_0)[\/latex]. The point [latex](x_0, y_0)[\/latex] is called a <strong>critical point<\/strong> of [latex]f[\/latex] if one of the following conditions holds:\r\n<ol>\r\n \t<li>[latex]f_x=(x_0, y_0)=f_y(x_0, y_0)=0[\/latex]<\/li>\r\n \t<li>Either\u00a0[latex]f_x(x_0, y_0)[\/latex] or\u00a0[latex]f_y(x_0, y_0)[\/latex] does not exist.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Find the critical points of each of the following functions:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]f(x,y)=\\sqrt{4y^2-9x^2+24y+36x+36}[\/latex]<\/li>\r\n \t<li>[latex]g(x,y)=x^2+2xy-4y^2+4x-6y+4[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"689124653\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689124653\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>First, we calculate [latex]f_x(x, y)[\/latex] and\u00a0[latex]f_y(x, y)[\/latex]:[latex]\\hspace{3cm}\\begin{align}\r\n\r\nf_x(x,y)&amp;=\\frac12(-18x+36)(4y^2-9x^2+24y+36x+36)^{-1\/2} \\\\\r\n\r\n&amp;=\\frac{-9x+18}{\\sqrt{4y^2-9x^2+24y+36x+36}} \\\\\r\n\r\nf_y(x,y)&amp;=\\frac12(8y+24)(4y^2-9x^2+24y+36x+36)^{-1\/2} \\\\\r\n\r\n&amp;=\\frac{4y+12}{\\sqrt{4y^2-9x^2+24y+36x+36}}.\r\n\r\n\\end{align}[\/latex]\r\n\r\nNext, we set\u00a0each of these expressions equal to zero:\r\n\r\n[latex]\\hspace{6cm}\\begin{align}\r\n\r\n\\frac{-9x+18}{\\sqrt{4y^2-9x^2+24y+36x+36}}&amp;=0 \\\\\r\n\r\n\\frac{4y+12}{\\sqrt{4y^2-9x^2+24y+36x+36}}&amp;=0.\r\n\r\n\\end{align}[\/latex]\r\n\r\nThen, multiply each equation by its common denominator:\r\n\r\n[latex]\\hspace{10cm}\\large{\\begin{align}\r\n\r\n-9x+18&amp;=0 \\\\\r\n\r\n4y+2&amp;=0.\r\n\r\n\\end{align}}[\/latex]\r\n\r\nTherefore,\u00a0[latex]x=2[\/latex] and\u00a0[latex]y=-3[\/latex], so\u00a0[latex](2, -3)[\/latex] is a critical point of\u00a0[latex]f[\/latex].\r\n\r\nWe must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:\r\n<p style=\"text-align: center;\">[latex]\\large{4y^{2}-9x^{2}+24y+36x+36=0}[\/latex]<\/p>\r\nThis equation represents a hyperbola. We should also note that the domain of [latex]f[\/latex] consists of points satisfying the inequality\r\n<p style=\"text-align: center;\">[latex]\\large{4y^{2}-9x^{2}+24y+36x+36\\geq 0}[\/latex]<\/p>\r\nTherefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:\r\n\r\n[latex]\\hspace{6cm}\\begin{align}\r\n\r\n4y^{2}-9x^{2}+24y+36x+36&amp;=0 \\\\\r\n\r\n4y^{2}-9x^{2}+24y+36x&amp;=-36 \\\\\r\n\r\n4y^{2}+24y-9x^{2}+36x&amp;=-36 \\\\\r\n\r\n4(y^2+6y)-9(x^2-4x)&amp;=-36 \\\\\r\n\r\n4(y^2+6y+9)-9(x^2-4x+4)&amp;=-36+36-36 \\\\\r\n\r\n4(y+3)^2-9(x-2)^2&amp;=-36.\r\n\r\n\\end{align}[\/latex]\r\n\r\nDividing both sides by\u00a0[latex]-36[\/latex] puts the equation in standard form:\r\n\r\n[latex]\\hspace{7.65cm}\\begin{align}\r\n\r\n\\frac{4(y+3)^2}{-36}-\\frac{9(x-2)^2}{-36}&amp;=1 \\\\\r\n\r\n\\frac{(y-2)^2}{4}-\\frac{(x+3)^2}{9}&amp;=1.\r\n\r\n\\end{align}[\/latex]\r\n\r\nNotice that point [latex](2, -3)[\/latex]\u00a0is the center of the hyperbola.<\/li>\r\n \t<li>First, we calculate\u00a0[latex]g_x(x, y)[\/latex] and\u00a0[latex]g_y(x, y)[\/latex]:[latex]\\hspace{7.5cm}\\large{\\begin{align}\r\n\r\ng_x(x,y)&amp;=2x+2y+4 \\\\\r\n\r\ng_y(x,y)&amp;=2x-8y-6.\r\n\r\n\\end{align}}[\/latex]\r\n\r\nNext, we set each of these expressions equal to zero, which gives a system of equations in\u00a0[latex]x[\/latex] and\u00a0[latex]y[\/latex]:\r\n\r\n[latex]\\hspace{8cm}\\large{\\begin{align}\r\n\r\n2x+2y+4&amp;=0 \\\\\r\n\r\n2x-8y-6&amp;=0.\r\n\r\n\\end{align}}[\/latex]\r\n\r\nSubtracting the second equation from the first gives [latex]10y+10=0[\/latex], so\u00a0[latex]y=-1[\/latex]. Substituting this into the first equation gives[latex]2x+2(-1)+4=0[\/latex], so [latex]x=-1[\/latex]. Therefore [latex](-1,-1)[\/latex] is a critical point of [latex]g[\/latex] (Figure 1). There are no points in [latex]\\mathbb{R}^2[\/latex] that make either partial derivative not exist.\r\n\r\n[caption id=\"attachment_1276\" align=\"aligncenter\" width=\"760\"]<img class=\"size-full wp-image-1276\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134045\/4-7-1.jpeg\" alt=\"The function g(x, y) = x2 + 2xy \u2013 4y2 + 4x \u2013 6y + 4 is shown with critical point (\u20131, \u20131, 5). The critical point is located where the derivative in the x and y directions are both zero.\" width=\"760\" height=\"418\" \/> Figure 1. The function [latex]\\small{g(x,y)}[\/latex] has a critical point at [latex]\\small{(-1,-1,5)}[\/latex].[\/caption]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3>Local Extrema<\/h3>\r\n<p class=\"whitespace-normal break-words\">The main purpose for determining critical points is to locate relative maxima and minima, just as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all other function values in that interval. For functions of two or more variables, we work with an open disk around the point.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>local extrema<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]z=f(x, y)[\/latex] be a function of two variables that is defined and continuous on an open set containing the point [latex](x_0, y_0)[\/latex].<\/p>\r\n[latex]\\\\[\/latex]\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Local Maximum:<\/strong> The function [latex]f[\/latex] has a <strong>local maximum<\/strong> at [latex](x_0, y_0)[\/latex] if\r\n<center>[latex]f(x_0, y_0)\\geq f(x, y)[\/latex]<\/center>\r\nfor all points [latex](x, y)[\/latex] within some disk centered at [latex](x_0, y_0)[\/latex].\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\">The number [latex]f(x_0, y_0)[\/latex] is called a <strong>local maximum value<\/strong>. If this inequality holds for every point [latex](x, y)[\/latex] in the domain of [latex]f[\/latex], then [latex]f[\/latex] has a <strong>global maximum<\/strong> (also called an <strong>absolute maximum<\/strong>) at [latex](x_0, y_0)[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Local Minimum:<\/strong> The function [latex]f[\/latex] has a <strong>local minimum<\/strong> at [latex](x_0, y_0)[\/latex] if\r\n<center>[latex]f(x_0, y_0)\\leq f(x, y)[\/latex]<\/center>\r\nfor all points [latex](x, y)[\/latex] within some disk centered at [latex](x_0, y_0)[\/latex].\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\">The number [latex]f(x_0, y_0)[\/latex] is called a <strong>local minimum value<\/strong>. If this inequality holds for every point [latex](x, y)[\/latex] in the domain of [latex]f[\/latex], then [latex]f[\/latex] has a <strong>global minimum<\/strong> (also called an <strong>absolute minimum<\/strong>) at [latex](x_0, y_0)[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">If [latex]f(x_0, y_0)[\/latex] is either a local maximum or local minimum value, then it is called a <strong>local extremum<\/strong>.<\/p>\r\n\r\n<\/section>[caption id=\"attachment_1278\" align=\"aligncenter\" width=\"479\"]<img class=\"size-full wp-image-1278\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134205\/4-7-2.jpeg\" alt=\"The function z = the square root of (16 \u2013 x2 \u2013 y2) is shown, which is the upper hemisphere of radius 4 with center at the origin. In the xy plane, the circle with radius 4 and center at the origin is highlighted; it has equation x2 + y2 = 16.\" width=\"479\" height=\"445\" \/> Figure 2. The graph of [latex]\\small{z=\\sqrt{16-x^{2}-y^{2}}}[\/latex] has a maximum value when [latex]\\small{(x,y)=(0,0)}[\/latex] It attains its minimum value at the boundary of its domain, which is the circle [latex]\\small{x^{2}+y^{2}=16}[\/latex].[\/caption]Just as with functions of one variable, extrema of functions of more than one variable occur at critical points.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>Fermat's theorem for functions of two variables<\/h3>\r\nLet [latex]z=f(x, y)[\/latex] be a function of two variables that is defined and continuous on an open set containing the point [latex](x_0, y_0)[\/latex]. Suppose [latex]f_x[\/latex] and [latex]f_y[\/latex] each exists at [latex](x_0, y_0)[\/latex]. If [latex]f[\/latex] has a local extremum at [latex](x_0, y_0)[\/latex], then [latex](x_0, y_0)[\/latex] is a critical point of [latex]f[\/latex].\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find critical points using partial derivatives<\/li>\n<li>Apply the second derivative test<\/li>\n<li>Find absolute extrema on closed regions<\/li>\n<\/ul>\n<\/section>\n<h2>Critical Points<\/h2>\n<p class=\"whitespace-normal break-words\">One of the most useful applications of derivatives for single-variable functions is finding maximum and minimum values. This application is equally important for functions of two or more variables. The main ideas\u2014finding critical points and using derivative tests\u2014are still valid, but more independent variables lead to more possible outcomes.<\/p>\n<p class=\"whitespace-normal break-words\">For functions of a single variable, critical points occur where the derivative equals zero or does not exist. For functions of two or more variables, the concept is the same, except we work with partial derivatives.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>critical point<\/h3>\n<p>Let [latex]z=f(x, y)[\/latex] be a function of two variables that is defined on an open set containing the point [latex](x_0, y_0)[\/latex]. The point [latex](x_0, y_0)[\/latex] is called a <strong>critical point<\/strong> of [latex]f[\/latex] if one of the following conditions holds:<\/p>\n<ol>\n<li>[latex]f_x=(x_0, y_0)=f_y(x_0, y_0)=0[\/latex]<\/li>\n<li>Either\u00a0[latex]f_x(x_0, y_0)[\/latex] or\u00a0[latex]f_y(x_0, y_0)[\/latex] does not exist.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find the critical points of each of the following functions:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]f(x,y)=\\sqrt{4y^2-9x^2+24y+36x+36}[\/latex]<\/li>\n<li>[latex]g(x,y)=x^2+2xy-4y^2+4x-6y+4[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q689124653\">Show Solution<\/button><\/p>\n<div id=\"q689124653\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>First, we calculate [latex]f_x(x, y)[\/latex] and\u00a0[latex]f_y(x, y)[\/latex]:[latex]\\hspace{3cm}\\begin{align}    f_x(x,y)&=\\frac12(-18x+36)(4y^2-9x^2+24y+36x+36)^{-1\/2} \\\\    &=\\frac{-9x+18}{\\sqrt{4y^2-9x^2+24y+36x+36}} \\\\    f_y(x,y)&=\\frac12(8y+24)(4y^2-9x^2+24y+36x+36)^{-1\/2} \\\\    &=\\frac{4y+12}{\\sqrt{4y^2-9x^2+24y+36x+36}}.    \\end{align}[\/latex]\n<p>Next, we set\u00a0each of these expressions equal to zero:<\/p>\n<p>[latex]\\hspace{6cm}\\begin{align}    \\frac{-9x+18}{\\sqrt{4y^2-9x^2+24y+36x+36}}&=0 \\\\    \\frac{4y+12}{\\sqrt{4y^2-9x^2+24y+36x+36}}&=0.    \\end{align}[\/latex]<\/p>\n<p>Then, multiply each equation by its common denominator:<\/p>\n<p>[latex]\\hspace{10cm}\\large{\\begin{align}    -9x+18&=0 \\\\    4y+2&=0.    \\end{align}}[\/latex]<\/p>\n<p>Therefore,\u00a0[latex]x=2[\/latex] and\u00a0[latex]y=-3[\/latex], so\u00a0[latex](2, -3)[\/latex] is a critical point of\u00a0[latex]f[\/latex].<\/p>\n<p>We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{4y^{2}-9x^{2}+24y+36x+36=0}[\/latex]<\/p>\n<p>This equation represents a hyperbola. We should also note that the domain of [latex]f[\/latex] consists of points satisfying the inequality<\/p>\n<p style=\"text-align: center;\">[latex]\\large{4y^{2}-9x^{2}+24y+36x+36\\geq 0}[\/latex]<\/p>\n<p>Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:<\/p>\n<p>[latex]\\hspace{6cm}\\begin{align}    4y^{2}-9x^{2}+24y+36x+36&=0 \\\\    4y^{2}-9x^{2}+24y+36x&=-36 \\\\    4y^{2}+24y-9x^{2}+36x&=-36 \\\\    4(y^2+6y)-9(x^2-4x)&=-36 \\\\    4(y^2+6y+9)-9(x^2-4x+4)&=-36+36-36 \\\\    4(y+3)^2-9(x-2)^2&=-36.    \\end{align}[\/latex]<\/p>\n<p>Dividing both sides by\u00a0[latex]-36[\/latex] puts the equation in standard form:<\/p>\n<p>[latex]\\hspace{7.65cm}\\begin{align}    \\frac{4(y+3)^2}{-36}-\\frac{9(x-2)^2}{-36}&=1 \\\\    \\frac{(y-2)^2}{4}-\\frac{(x+3)^2}{9}&=1.    \\end{align}[\/latex]<\/p>\n<p>Notice that point [latex](2, -3)[\/latex]\u00a0is the center of the hyperbola.<\/li>\n<li>First, we calculate\u00a0[latex]g_x(x, y)[\/latex] and\u00a0[latex]g_y(x, y)[\/latex]:[latex]\\hspace{7.5cm}\\large{\\begin{align}    g_x(x,y)&=2x+2y+4 \\\\    g_y(x,y)&=2x-8y-6.    \\end{align}}[\/latex]\n<p>Next, we set each of these expressions equal to zero, which gives a system of equations in\u00a0[latex]x[\/latex] and\u00a0[latex]y[\/latex]:<\/p>\n<p>[latex]\\hspace{8cm}\\large{\\begin{align}    2x+2y+4&=0 \\\\    2x-8y-6&=0.    \\end{align}}[\/latex]<\/p>\n<p>Subtracting the second equation from the first gives [latex]10y+10=0[\/latex], so\u00a0[latex]y=-1[\/latex]. Substituting this into the first equation gives[latex]2x+2(-1)+4=0[\/latex], so [latex]x=-1[\/latex]. Therefore [latex](-1,-1)[\/latex] is a critical point of [latex]g[\/latex] (Figure 1). There are no points in [latex]\\mathbb{R}^2[\/latex] that make either partial derivative not exist.<\/p>\n<figure id=\"attachment_1276\" aria-describedby=\"caption-attachment-1276\" style=\"width: 760px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1276\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134045\/4-7-1.jpeg\" alt=\"The function g(x, y) = x2 + 2xy \u2013 4y2 + 4x \u2013 6y + 4 is shown with critical point (\u20131, \u20131, 5). The critical point is located where the derivative in the x and y directions are both zero.\" width=\"760\" height=\"418\" \/><figcaption id=\"caption-attachment-1276\" class=\"wp-caption-text\">Figure 1. The function [latex]\\small{g(x,y)}[\/latex] has a critical point at [latex]\\small{(-1,-1,5)}[\/latex].<\/figcaption><\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<h3>Local Extrema<\/h3>\n<p class=\"whitespace-normal break-words\">The main purpose for determining critical points is to locate relative maxima and minima, just as in single-variable calculus. When working with a function of one variable, the definition of a local extremum involves finding an interval around the critical point such that the function value is either greater than or less than all other function values in that interval. For functions of two or more variables, we work with an open disk around the point.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>local extrema<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]z=f(x, y)[\/latex] be a function of two variables that is defined and continuous on an open set containing the point [latex](x_0, y_0)[\/latex].<\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Local Maximum:<\/strong> The function [latex]f[\/latex] has a <strong>local maximum<\/strong> at [latex](x_0, y_0)[\/latex] if\n<div style=\"text-align: center;\">[latex]f(x_0, y_0)\\geq f(x, y)[\/latex]<\/div>\n<p>for all points [latex](x, y)[\/latex] within some disk centered at [latex](x_0, y_0)[\/latex].<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">The number [latex]f(x_0, y_0)[\/latex] is called a <strong>local maximum value<\/strong>. If this inequality holds for every point [latex](x, y)[\/latex] in the domain of [latex]f[\/latex], then [latex]f[\/latex] has a <strong>global maximum<\/strong> (also called an <strong>absolute maximum<\/strong>) at [latex](x_0, y_0)[\/latex].<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Local Minimum:<\/strong> The function [latex]f[\/latex] has a <strong>local minimum<\/strong> at [latex](x_0, y_0)[\/latex] if\n<div style=\"text-align: center;\">[latex]f(x_0, y_0)\\leq f(x, y)[\/latex]<\/div>\n<p>for all points [latex](x, y)[\/latex] within some disk centered at [latex](x_0, y_0)[\/latex].<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">The number [latex]f(x_0, y_0)[\/latex] is called a <strong>local minimum value<\/strong>. If this inequality holds for every point [latex](x, y)[\/latex] in the domain of [latex]f[\/latex], then [latex]f[\/latex] has a <strong>global minimum<\/strong> (also called an <strong>absolute minimum<\/strong>) at [latex](x_0, y_0)[\/latex].<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">If [latex]f(x_0, y_0)[\/latex] is either a local maximum or local minimum value, then it is called a <strong>local extremum<\/strong>.<\/p>\n<\/section>\n<figure id=\"attachment_1278\" aria-describedby=\"caption-attachment-1278\" style=\"width: 479px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1278\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/09\/23134205\/4-7-2.jpeg\" alt=\"The function z = the square root of (16 \u2013 x2 \u2013 y2) is shown, which is the upper hemisphere of radius 4 with center at the origin. In the xy plane, the circle with radius 4 and center at the origin is highlighted; it has equation x2 + y2 = 16.\" width=\"479\" height=\"445\" \/><figcaption id=\"caption-attachment-1278\" class=\"wp-caption-text\">Figure 2. The graph of [latex]\\small{z=\\sqrt{16-x^{2}-y^{2}}}[\/latex] has a maximum value when [latex]\\small{(x,y)=(0,0)}[\/latex] It attains its minimum value at the boundary of its domain, which is the circle [latex]\\small{x^{2}+y^{2}=16}[\/latex].<\/figcaption><\/figure>\n<p>Just as with functions of one variable, extrema of functions of more than one variable occur at critical points.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>Fermat&#8217;s theorem for functions of two variables<\/h3>\n<p>Let [latex]z=f(x, y)[\/latex] be a function of two variables that is defined and continuous on an open set containing the point [latex](x_0, y_0)[\/latex]. Suppose [latex]f_x[\/latex] and [latex]f_y[\/latex] each exists at [latex](x_0, y_0)[\/latex]. If [latex]f[\/latex] has a local extremum at [latex](x_0, y_0)[\/latex], then [latex](x_0, y_0)[\/latex] is a critical point of [latex]f[\/latex].<\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":4,"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-652","chapter","type-chapter","status-publish","hentry"],"part":430,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/652","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/652\/revisions"}],"predecessor-version":[{"id":1827,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/652\/revisions\/1827"}],"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\/652\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=652"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=652"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=652"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=652"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}