{"id":3934,"date":"2022-04-05T18:52:09","date_gmt":"2022-04-05T18:52:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3934"},"modified":"2022-10-29T02:14:42","modified_gmt":"2022-10-29T02:14:42","slug":"critical-points","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/critical-points\/","title":{"raw":"Critical Points","rendered":"Critical Points"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li><span class=\"os-abstract-content\">Use partial derivatives to locate critical points for a function of two variables.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167793543824\">For functions of a single variable, we defined critical points as the values of the function when the derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]z=f(x, y)[\/latex] be a function of\u00a0two 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\u00a0<span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term195\" data-type=\"term\"><strong>critical point of a function of two variables<\/strong> [latex]f[\/latex]\u00a0<\/span>if one of the two 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<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding critical points<\/h3>\r\nFind the critical points of each of the following functions:\r\n\r\na.\u00a0[latex]f(x,y)=\\sqrt{4y^2-9x^2+24y+36x+36}[\/latex]\r\n\r\nb.\u00a0[latex]g(x,y)=x^2+2xy-4y^2+4x-6y+4[\/latex]\r\n\r\n[reveal-answer q=\"689124653\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689124653\"]\r\n\r\na. First, we calculate [latex]f_x(x, y)[\/latex] and\u00a0[latex]f_y(x, y)[\/latex]:\r\n\r\n[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.\r\n\r\nb. First, we calculate\u00a0[latex]g_x(x, y)[\/latex] and\u00a0[latex]g_y(x, y)[\/latex]:\r\n\r\n[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][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the critical point of the function\u00a0[latex]f(x, y)=x^{3}+2xy-2x-4y[\/latex].\r\n\r\n[reveal-answer q=\"689124632\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689124632\"]\r\n\r\n[latex](2, -5)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186167&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Q5SSTSx5i1o&amp;video_target=tpm-plugin-w2org5v0-Q5SSTSx5i1o\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.34_transcript.html\">transcript for \u201cCP 4.34\u201d here (opens in new window).<\/a><\/center>The main purpose for determining critical points is to locate relative maxima and minima, 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 the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794063817\">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]. Then [latex]f[\/latex] has a\u00a0<em data-effect=\"italics\">local maximum<\/em>\u00a0at [latex](x_0, y_0)[\/latex] if<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">[latex]f(x_0, y_0)\\geq f(x, y)[\/latex]<\/span>\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">for all points [latex](x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">within some disk centered at [latex](x_0, y_0)[\/latex].\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">The number [latex]f(x_0, y_0)[\/latex] is<\/span><span style=\"font-size: 1rem; text-align: initial;\">\u00a0called a\u00a0<\/span><em style=\"font-size: 1rem; text-align: initial;\" data-effect=\"italics\">local maximum value<\/em><span style=\"font-size: 1rem; text-align: initial;\">. If the preceding inequality holds for every point [latex](x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">in the domain of [latex]f[\/latex],\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">then [latex]f[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">has a\u00a0<\/span><em style=\"font-size: 1rem; text-align: initial;\" data-effect=\"italics\">global maximum<\/em><span style=\"font-size: 1rem; text-align: initial;\">\u00a0(also called an\u00a0<\/span><em style=\"font-size: 1rem; text-align: initial;\" data-effect=\"italics\">absolute maximum<\/em><span style=\"font-size: 1rem; text-align: initial;\">) at [latex](x_0, y_0)[\/latex].<\/span>\r\n\r\nThe function [latex]f[\/latex] has a\u00a0<em data-effect=\"italics\">local minimum<\/em>\u00a0at [latex](x_0, y_0)[\/latex] if\r\n\r\n[latex]f(x_0, y_0)\\leq f(x, y)[\/latex]\r\n\r\nfor all points [latex](x, y)[\/latex]\u00a0within some disk centered at [latex](x_0, y_0)[\/latex]. The number [latex]f(x_0, y_0)[\/latex] is called a\u00a0<em data-effect=\"italics\">local minimum value<\/em>. If the preceding inequality holds for every point [latex](x, y)[\/latex]\u00a0in the domain of [latex]f[\/latex], then [latex]f[\/latex] has a\u00a0<em data-effect=\"italics\">global minimum<\/em>\u00a0(also called an\u00a0<em data-effect=\"italics\">absolute minimum<\/em>) at [latex](x_0, y_0)[\/latex].\r\n\r\nIf [latex]f(x_0, y_0)[\/latex]\u00a0is either a local maximum or local minimum value, then it is called a\u00a0<span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term196\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">local extremum<\/em><\/span>\u00a0(see the following figure).\r\n\r\n<\/div>\r\n[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]In\u00a0Maxima and Minima, we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem: Fermat's theorem for functions of two variables<\/h3>\r\n\r\n<hr \/>\r\n\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\u00a0[latex]f[\/latex] has a\u00a0local extremum\u00a0at\u00a0[latex](x_0, y_0)[\/latex], then\u00a0[latex](x_0, y_0)[\/latex] is a critical point of\u00a0[latex]f[\/latex].\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li><span class=\"os-abstract-content\">Use partial derivatives to locate critical points for a function of two variables.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167793543824\">For functions of a single variable, we defined critical points as the values of the function when the derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p>Let [latex]z=f(x, y)[\/latex] be a function of\u00a0two 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\u00a0<span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term195\" data-type=\"term\"><strong>critical point of a function of two variables<\/strong> [latex]f[\/latex]\u00a0<\/span>if one of the two 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<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding critical points<\/h3>\n<p>Find the critical points of each of the following functions:<\/p>\n<p>a.\u00a0[latex]f(x,y)=\\sqrt{4y^2-9x^2+24y+36x+36}[\/latex]<\/p>\n<p>b.\u00a0[latex]g(x,y)=x^2+2xy-4y^2+4x-6y+4[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q689124653\">Show Solution<\/span><\/p>\n<div id=\"q689124653\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. First, we calculate [latex]f_x(x, y)[\/latex] and\u00a0[latex]f_y(x, y)[\/latex]:<\/p>\n<p>[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]<\/p>\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.<\/p>\n<p>b. First, we calculate\u00a0[latex]g_x(x, y)[\/latex] and\u00a0[latex]g_y(x, y)[\/latex]:<\/p>\n<p>[latex]\\hspace{7.5cm}\\large{\\begin{align}    g_x(x,y)&=2x+2y+4 \\\\    g_y(x,y)&=2x-8y-6.    \\end{align}}[\/latex]<\/p>\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<div id=\"attachment_1276\" style=\"width: 770px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1276\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the critical point of the function\u00a0[latex]f(x, y)=x^{3}+2xy-2x-4y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q689124632\">Show Solution<\/span><\/p>\n<div id=\"q689124632\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](2, -5)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186167&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Q5SSTSx5i1o&amp;video_target=tpm-plugin-w2org5v0-Q5SSTSx5i1o\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.34_transcript.html\">transcript for \u201cCP 4.34\u201d here (opens in new window).<\/a><\/div>\n<p>The main purpose for determining critical points is to locate relative maxima and minima, 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 the other function values in that interval. When working with a function of two or more variables, we work with an open disk around the point.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167794063817\">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]. Then [latex]f[\/latex] has a\u00a0<em data-effect=\"italics\">local maximum<\/em>\u00a0at [latex](x_0, y_0)[\/latex] if<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">[latex]f(x_0, y_0)\\geq f(x, y)[\/latex]<\/span><\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">for all points [latex](x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">within some disk centered at [latex](x_0, y_0)[\/latex].\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">The number [latex]f(x_0, y_0)[\/latex] is<\/span><span style=\"font-size: 1rem; text-align: initial;\">\u00a0called a\u00a0<\/span><em style=\"font-size: 1rem; text-align: initial;\" data-effect=\"italics\">local maximum value<\/em><span style=\"font-size: 1rem; text-align: initial;\">. If the preceding inequality holds for every point [latex](x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">in the domain of [latex]f[\/latex],\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">then [latex]f[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">has a\u00a0<\/span><em style=\"font-size: 1rem; text-align: initial;\" data-effect=\"italics\">global maximum<\/em><span style=\"font-size: 1rem; text-align: initial;\">\u00a0(also called an\u00a0<\/span><em style=\"font-size: 1rem; text-align: initial;\" data-effect=\"italics\">absolute maximum<\/em><span style=\"font-size: 1rem; text-align: initial;\">) at [latex](x_0, y_0)[\/latex].<\/span><\/p>\n<p>The function [latex]f[\/latex] has a\u00a0<em data-effect=\"italics\">local minimum<\/em>\u00a0at [latex](x_0, y_0)[\/latex] if<\/p>\n<p>[latex]f(x_0, y_0)\\leq f(x, y)[\/latex]<\/p>\n<p>for all points [latex](x, y)[\/latex]\u00a0within some disk centered at [latex](x_0, y_0)[\/latex]. The number [latex]f(x_0, y_0)[\/latex] is called a\u00a0<em data-effect=\"italics\">local minimum value<\/em>. If the preceding inequality holds for every point [latex](x, y)[\/latex]\u00a0in the domain of [latex]f[\/latex], then [latex]f[\/latex] has a\u00a0<em data-effect=\"italics\">global minimum<\/em>\u00a0(also called an\u00a0<em data-effect=\"italics\">absolute minimum<\/em>) at [latex](x_0, y_0)[\/latex].<\/p>\n<p>If [latex]f(x_0, y_0)[\/latex]\u00a0is either a local maximum or local minimum value, then it is called a\u00a0<span id=\"7fafffc3-233c-4832-ade9-1ca1f91ec9c4_term196\" class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">local extremum<\/em><\/span>\u00a0(see the following figure).<\/p>\n<\/div>\n<div id=\"attachment_1278\" style=\"width: 489px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1278\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<p>In\u00a0Maxima and Minima, we showed that extrema of functions of one variable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem: Fermat&#8217;s theorem for functions of two variables<\/h3>\n<hr \/>\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\u00a0[latex]f[\/latex] has a\u00a0local extremum\u00a0at\u00a0[latex](x_0, y_0)[\/latex], then\u00a0[latex](x_0, y_0)[\/latex] is a critical point of\u00a0[latex]f[\/latex].<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3934\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 4.34. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":30,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 4.34\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3934","chapter","type-chapter","status-publish","hentry"],"part":22,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3934","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3934\/revisions"}],"predecessor-version":[{"id":5918,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3934\/revisions\/5918"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/22"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/3934\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=3934"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=3934"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=3934"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=3934"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}