{"id":654,"date":"2025-10-09T15:28:18","date_gmt":"2025-10-09T15:28:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=654"},"modified":"2025-10-29T14:36:59","modified_gmt":"2025-10-29T14:36:59","slug":"critical-points-and-local-extrema-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/critical-points-and-local-extrema-fresh-take\/","title":{"raw":"Critical Points and Local Extrema: Fresh Take","rendered":"Critical Points and Local Extrema: Fresh Take"},"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<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Finding where a function reaches its highest or lowest values is one of the most practical applications of calculus. For functions of two variables, the process mirrors single-variable calculus but requires checking both partial derivatives.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>What Are Critical Points?<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A point [latex](x_0, y_0)[\/latex] is a critical point of [latex]f(x, y)[\/latex] if:<\/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\">Both partial derivatives are zero: [latex]f_x(x_0, y_0) = 0[\/latex] and [latex]f_y(x_0, y_0) = 0[\/latex], OR<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Either [latex]f_x(x_0, y_0)[\/latex] or [latex]f_y(x_0, y_0)[\/latex] doesn't exist<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Critical points are candidates for local maxima and minima.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Local Extrema<\/strong><\/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\">A <strong>local maximum<\/strong> at [latex](x_0, y_0)[\/latex] means [latex]f(x_0, y_0)[\/latex] is greater than or equal to all nearby function values<\/li>\r\n \t<li class=\"whitespace-normal break-words\">A <strong>local minimum<\/strong> at [latex](x_0, y_0)[\/latex] means [latex]f(x_0, y_0)[\/latex] is less than or equal to all nearby function values<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If the inequality holds everywhere in the domain, you have a <strong>global (absolute)<\/strong> maximum or minimum<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/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>Calculate both partial derivatives<\/strong> [latex]f_x[\/latex] and [latex]f_y[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Set both equal to zero<\/strong> and solve the system of equations<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Check where partials don't exist<\/strong> (look for division by zero, square roots of negatives, etc.)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Verify your solutions<\/strong> by substituting back into the original equations<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Insight from Fermat's Theorem<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]f[\/latex] has a local extremum at [latex](x_0, y_0)[\/latex] and both partial derivatives exist there, then [latex](x_0, y_0)[\/latex] must be a critical point. This tells you where to look for extrema, but not every critical point is an extremum\u2014you'll need additional tests to determine the nature of each critical point.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the critical point of the function\u00a0[latex]f(x, y)=x^{3}+2xy-2x-4y[\/latex].[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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section>\r\n<h2>Second Derivative Test<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Once you've found critical points, you need to determine whether each one is a local maximum, local minimum, or saddle point. The second derivative test gives you a systematic way to classify them using second-order partial derivatives.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Discriminant Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">At a critical point [latex](x_0, y_0)[\/latex], calculate:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]D = f_{xx}(x_0, y_0)f_{yy}(x_0, y_0) - (f_{xy}(x_0, y_0))^2[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This discriminant [latex]D[\/latex] replaces the single second derivative from single-variable calculus.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Interpreting the Test<\/strong><\/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\"><strong>If [latex]D &gt; 0[\/latex] and [latex]f_{xx} &gt; 0[\/latex]<\/strong>: Local minimum (bowl-shaped, curves up)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>If [latex]D &gt; 0[\/latex] and [latex]f_{xx} &lt; 0[\/latex]<\/strong>: Local maximum (dome-shaped, curves down)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>If [latex]D &lt; 0[\/latex]<\/strong>: Saddle point (curves up in one direction, down in another)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>If [latex]D = 0[\/latex]<\/strong>: Test is inconclusive (need other methods)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>What's a Saddle Point?<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A saddle point is where both partial derivatives are zero but the point is neither a maximum nor minimum. Picture a horse saddle: it curves up in one direction and down in another. The point [latex](0, 0)[\/latex] on [latex]f(x, y) = x^2 - y^2[\/latex] is a classic example.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/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 all critical points<\/strong> by solving [latex]f_x = 0[\/latex] and [latex]f_y = 0[\/latex] simultaneously<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Calculate all three second partial derivatives<\/strong>: [latex]f_{xx}[\/latex], [latex]f_{yy}[\/latex], and [latex]f_{xy}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Evaluate [latex]D[\/latex]<\/strong> at each critical point<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Apply the test<\/strong> to classify each point<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Reminder<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When [latex]D &gt; 0[\/latex], check the sign of [latex]f_{xx}[\/latex] (or [latex]f_{yy}[\/latex]\u2014either works). Positive means minimum, negative means maximum.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Use the second derivative to find the local extrema of the function[latex]f(x, y)=x^{3}+2xy-6x-4y^{2}[\/latex].\r\n\r\n[reveal-answer q=\"233578991\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"233578991\"]\r\n\r\n[latex](\\frac{4}{3},\\frac{1}{3})[\/latex] is a saddle point,\u00a0[latex](-\\frac{3}{2},-\\frac{3}{8})[\/latex] is a local maximum.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186168&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=0IESE6fWxpQ&amp;video_target=tpm-plugin-0hltm37e-0IESE6fWxpQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.35_transcript.html\">transcript for \u201cCP 4.35\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>Absolute Maxima and Minima<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Finding absolute (global) extrema on a closed, bounded region requires checking both the interior and the boundary of your domain. Unlike local extrema, absolute extrema compare function values across the entire region.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Extreme Value Theorem<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A continuous function on a closed, bounded set is guaranteed to have both an absolute maximum and an absolute minimum somewhere in that region. Your job is to find where they occur.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Where to Look<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Absolute extrema can only occur in two places:<\/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\"><strong>Interior critical points<\/strong>: where both [latex]f_x = 0[\/latex] and [latex]f_y = 0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Points on the boundary<\/strong>: the edge of your domain<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/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 interior critical points<\/strong>: Solve [latex]f_x = 0[\/latex] and [latex]f_y = 0[\/latex] simultaneously, keeping only points inside your domain<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Evaluate [latex]f[\/latex]<\/strong> at each interior critical point<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Find extrema on the boundary<\/strong>: Parameterize the boundary and find maximum\/minimum values there (this is often the hardest step)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Compare all values<\/strong>: The largest is your absolute maximum, the smallest is your absolute minimum<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Handling the Boundary<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For rectangular domains, check each edge separately by treating one variable as constant. For circular boundaries, use parametric equations like [latex]x = r\\cos t[\/latex], [latex]y = r\\sin t[\/latex]. Don't forget to check corner points on rectangular domains\u2014they're often where extrema occur.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Insight<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The absolute extrema might not be at critical points. They frequently occur on the boundary, which is why checking the boundary thoroughly is essential. Compare every candidate value you find\u2014no shortcuts.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1167793930156\">Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function<\/p>\r\n<p style=\"text-align: center;\">[latex]f(x, y)=4x^{2}-2xy+6y^{2}-8x+2y+3[\/latex]<\/p>\r\non the domain defined by\u00a0[latex]0\\leq x\\leq 2[\/latex] and\u00a0[latex]-1\\leq y\\leq 3[\/latex].\r\n\r\n[reveal-answer q=\"245883100\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"245883100\"]\r\n<p id=\"fs-id1167793951496\">The absolute minimum occurs at [latex](1,0)[\/latex]: [latex]f(1, 0)=-1[\/latex]. The absolute maximum occurs at [latex](0, 3)[\/latex]: [latex]f(0, 3)=63[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8186169&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=F6MTQg1RjYQ&amp;video_target=tpm-plugin-xetvxykn-F6MTQg1RjYQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP4.36_transcript.html\">transcript for \u201cCP 4.36\u201d here (opens in new window).<\/a><\/center><\/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<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Finding where a function reaches its highest or lowest values is one of the most practical applications of calculus. For functions of two variables, the process mirrors single-variable calculus but requires checking both partial derivatives.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>What Are Critical Points?<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A point [latex](x_0, y_0)[\/latex] is a critical point of [latex]f(x, y)[\/latex] if:<\/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\">Both partial derivatives are zero: [latex]f_x(x_0, y_0) = 0[\/latex] and [latex]f_y(x_0, y_0) = 0[\/latex], OR<\/li>\n<li class=\"whitespace-normal break-words\">Either [latex]f_x(x_0, y_0)[\/latex] or [latex]f_y(x_0, y_0)[\/latex] doesn&#8217;t exist<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Critical points are candidates for local maxima and minima.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Local Extrema<\/strong><\/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\">A <strong>local maximum<\/strong> at [latex](x_0, y_0)[\/latex] means [latex]f(x_0, y_0)[\/latex] is greater than or equal to all nearby function values<\/li>\n<li class=\"whitespace-normal break-words\">A <strong>local minimum<\/strong> at [latex](x_0, y_0)[\/latex] means [latex]f(x_0, y_0)[\/latex] is less than or equal to all nearby function values<\/li>\n<li class=\"whitespace-normal break-words\">If the inequality holds everywhere in the domain, you have a <strong>global (absolute)<\/strong> maximum or minimum<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/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>Calculate both partial derivatives<\/strong> [latex]f_x[\/latex] and [latex]f_y[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Set both equal to zero<\/strong> and solve the system of equations<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Check where partials don&#8217;t exist<\/strong> (look for division by zero, square roots of negatives, etc.)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Verify your solutions<\/strong> by substituting back into the original equations<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Key Insight from Fermat&#8217;s Theorem<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If [latex]f[\/latex] has a local extremum at [latex](x_0, y_0)[\/latex] and both partial derivatives exist there, then [latex](x_0, y_0)[\/latex] must be a critical point. This tells you where to look for extrema, but not every critical point is an extremum\u2014you&#8217;ll need additional tests to determine the nature of each critical point.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q689124632\">Show Solution<\/button><\/p>\n<div id=\"q689124632\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](2, -5)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<h2>Second Derivative Test<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Once you&#8217;ve found critical points, you need to determine whether each one is a local maximum, local minimum, or saddle point. The second derivative test gives you a systematic way to classify them using second-order partial derivatives.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Discriminant Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">At a critical point [latex](x_0, y_0)[\/latex], calculate:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]D = f_{xx}(x_0, y_0)f_{yy}(x_0, y_0) - (f_{xy}(x_0, y_0))^2[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This discriminant [latex]D[\/latex] replaces the single second derivative from single-variable calculus.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Interpreting the Test<\/strong><\/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\"><strong>If [latex]D > 0[\/latex] and [latex]f_{xx} > 0[\/latex]<\/strong>: Local minimum (bowl-shaped, curves up)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>If [latex]D > 0[\/latex] and [latex]f_{xx} < 0[\/latex]<\/strong>: Local maximum (dome-shaped, curves down)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>If [latex]D < 0[\/latex]<\/strong>: Saddle point (curves up in one direction, down in another)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>If [latex]D = 0[\/latex]<\/strong>: Test is inconclusive (need other methods)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>What&#8217;s a Saddle Point?<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A saddle point is where both partial derivatives are zero but the point is neither a maximum nor minimum. Picture a horse saddle: it curves up in one direction and down in another. The point [latex](0, 0)[\/latex] on [latex]f(x, y) = x^2 - y^2[\/latex] is a classic example.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/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 all critical points<\/strong> by solving [latex]f_x = 0[\/latex] and [latex]f_y = 0[\/latex] simultaneously<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Calculate all three second partial derivatives<\/strong>: [latex]f_{xx}[\/latex], [latex]f_{yy}[\/latex], and [latex]f_{xy}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Evaluate [latex]D[\/latex]<\/strong> at each critical point<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Apply the test<\/strong> to classify each point<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Key Reminder<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When [latex]D > 0[\/latex], check the sign of [latex]f_{xx}[\/latex] (or [latex]f_{yy}[\/latex]\u2014either works). Positive means minimum, negative means maximum.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Use the second derivative to find the local extrema of the function[latex]f(x, y)=x^{3}+2xy-6x-4y^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q233578991\">Show Solution<\/button><\/p>\n<div id=\"q233578991\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](\\frac{4}{3},\\frac{1}{3})[\/latex] is a saddle point,\u00a0[latex](-\\frac{3}{2},-\\frac{3}{8})[\/latex] is a local maximum.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186168&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=0IESE6fWxpQ&amp;video_target=tpm-plugin-0hltm37e-0IESE6fWxpQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP4.35_transcript.html\">transcript for \u201cCP 4.35\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>Absolute Maxima and Minima<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Finding absolute (global) extrema on a closed, bounded region requires checking both the interior and the boundary of your domain. Unlike local extrema, absolute extrema compare function values across the entire region.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Extreme Value Theorem<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A continuous function on a closed, bounded set is guaranteed to have both an absolute maximum and an absolute minimum somewhere in that region. Your job is to find where they occur.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Where to Look<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Absolute extrema can only occur in two places:<\/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\"><strong>Interior critical points<\/strong>: where both [latex]f_x = 0[\/latex] and [latex]f_y = 0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Points on the boundary<\/strong>: the edge of your domain<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/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 interior critical points<\/strong>: Solve [latex]f_x = 0[\/latex] and [latex]f_y = 0[\/latex] simultaneously, keeping only points inside your domain<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Evaluate [latex]f[\/latex]<\/strong> at each interior critical point<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Find extrema on the boundary<\/strong>: Parameterize the boundary and find maximum\/minimum values there (this is often the hardest step)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Compare all values<\/strong>: The largest is your absolute maximum, the smallest is your absolute minimum<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Handling the Boundary<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For rectangular domains, check each edge separately by treating one variable as constant. For circular boundaries, use parametric equations like [latex]x = r\\cos t[\/latex], [latex]y = r\\sin t[\/latex]. Don&#8217;t forget to check corner points on rectangular domains\u2014they&#8217;re often where extrema occur.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key Insight<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The absolute extrema might not be at critical points. They frequently occur on the boundary, which is why checking the boundary thoroughly is essential. Compare every candidate value you find\u2014no shortcuts.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1167793930156\">Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extrema of the function<\/p>\n<p style=\"text-align: center;\">[latex]f(x, y)=4x^{2}-2xy+6y^{2}-8x+2y+3[\/latex]<\/p>\n<p>on the domain defined by\u00a0[latex]0\\leq x\\leq 2[\/latex] and\u00a0[latex]-1\\leq y\\leq 3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q245883100\">Show Solution<\/button><\/p>\n<div id=\"q245883100\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793951496\">The absolute minimum occurs at [latex](1,0)[\/latex]: [latex]f(1, 0)=-1[\/latex]. The absolute maximum occurs at [latex](0, 3)[\/latex]: [latex]f(0, 3)=63[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8186169&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=F6MTQg1RjYQ&amp;video_target=tpm-plugin-xetvxykn-F6MTQg1RjYQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP4.36_transcript.html\">transcript for \u201cCP 4.36\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":8,"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-654","chapter","type-chapter","status-publish","hentry"],"part":430,"module-header":"fresh_take","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\/654","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":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/654\/revisions"}],"predecessor-version":[{"id":1839,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/654\/revisions\/1839"}],"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\/654\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=654"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=654"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=654"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=654"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}