{"id":131848,"date":"2022-07-20T08:00:37","date_gmt":"2022-07-20T13:00:37","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=131848"},"modified":"2026-03-28T11:56:25","modified_gmt":"2026-03-28T16:56:25","slug":"viruses-and-antiviral-drugs","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/solving-optimization-problems\/","title":{"rendered":"Solving Optimization Problems"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_VCkDZVPkfEI\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_VCkDZVPkfEI\" data-source-videoID=\"VCkDZVPkfEI\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Solving Optimization Problems Video\" height=\"720\" width=\"1280\" class=\"size-full\" data-matomo-title = \"Solving Optimization Problems\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_VCkDZVPkfEI:hover {cursor:pointer;} img#videoThumbnailImage_VCkDZVPkfEI {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/optimization-thumb.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_VCkDZVPkfEI\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_VCkDZVPkfEI\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Solving Optimization Problems\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_VCkDZVPkfEI\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_VCkDZVPkfEI\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_VCkDZVPkfEI\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction QzI_Function() {\n  var x = document.getElementById(\"QzI\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"QzI_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"QzI\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Optimization_Problem_1\" class=\"smooth-scroll\">Optimization Problem #1<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Optimization_Problem_2\" class=\"smooth-scroll\">Optimization Problem #2<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>You have likely experienced a situation where you needed to find a balance between two or more variables, in order to achieve an optimal result. Solving these kinds of optimization problems is all about finding the measurements or circumstances that ensure the best, or most efficient, outcome.<\/p>\n<p>In this video, I\u2019m going to show you how to solve these kinds of problems. Let\u2019s get started!<\/p>\n<h2><span id=\"Optimization_Problem_1\" class=\"m-toc-anchor\"><\/span>Optimization Problem #1<\/h2>\n<p>\nSuppose a rancher wants to enclose a field on his property for livestock, and he has materials to build 400 meters of fencing. What should the enclosed field\u2019s length and width be in order to maximize the area of the field?<\/p>\n<p>Let\u2019s take a look at some dimension samples and their areas to get a feel for how changing the length and width relates to changes in the area. <\/p>\n<p>One way the rancher could build the fence is with the width of 50 meters and a length of 150 meters. This would use up all 400 meters of fencing.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-1.webp\" alt=\"Green rectangle with dimensions 150 m by 50 m. Below, a blue rectangle displays a perimeter calculation: 50 m + 150 m + 50 m + 150 m = 400 m.\" width=\"600.18\" height=\"425.04\" class=\"aligncenter size-full wp-image-237613\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-1.webp 1429w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-1-300x212.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-1-1024x725.webp 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-1-768x544.webp 768w\" sizes=\"(max-width: 1429px) 100vw, 1429px\" \/><\/p>\n<p>With this setup, the rancher\u2019s field would have an area of \\(50\\times 150=7,500\\) square meters.<\/p>\n<p>However, if the rancher chooses to increase the width to 60 meters and therefore decrease the length to 140 meters, the area of the field would then be 8,400 square meters, which is a sizeable improvement!<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-2.webp\" alt=\"A green rectangle labeled 140 m by 60 m with surrounding dimensions. Below, a calculation: 60 m \u00d7 140 m = 8,400 m\u00b2 on an orange rectangle.\" width=\"597.96\" height=\"418\" class=\"aligncenter size-full wp-image-237616\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-2.webp 1359w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-2-300x210.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-2-1024x716.webp 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-2-768x537.webp 768w\" sizes=\"(max-width: 1359px) 100vw, 1359px\" \/><\/p>\n<p>If we keep increasing the width and decreasing the length, we will continue to see the increases in area\u2026up to a point. Eventually, a maximum area will be reached and then the area will begin to decrease. For example, if we increase the width all the way up to 199 meters, and decrease the length accordingly to just 1 meter, the area of the field will only be 199 square meters.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-3.webp\" alt=\"Diagram showing a long vertical line labeled &quot;199 m&quot; and a short horizontal line labeled &quot;1 m,&quot; forming an L shape. A box displays the equation &quot;1 m \u00d7 199 m = 199 m\u00b2.\" width=\"527.99\" height=\"512.82\" class=\"aligncenter size-full wp-image-237619\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-3.webp 1427w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-3-300x291.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-3-1024x995.webp 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/rancher-exaxmple-3-768x746.webp 768w\" sizes=\"(max-width: 1427px) 100vw, 1427px\" \/><\/p>\n<p>As we can see, the area of the field very much depends on its dimensions, and there should be a \u201csweet spot,\u201d a set of dimensions which maximizes the area of the field. How can we determine exactly what those optimal dimensions are?<\/p>\n<p>The good news is you won\u2019t need to use trial and error for these problems.<\/p>\n<h3><span id=\"Critical_Points\" class=\"m-toc-anchor\"><\/span>Critical Points<\/h3>\n<p>\nBefore we solve this field problem, we need to talk about critical points.<\/p>\n<p>Consider this shape.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-138829\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/08\/I09882-Solving-Optimization-Problems-4-1-e1661523352688.png\" alt=\"Parabola with an exact point of 0\" width=\"400\" height=\"400\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/08\/I09882-Solving-Optimization-Problems-4-1-e1661523352688.png 400w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/08\/I09882-Solving-Optimization-Problems-4-1-e1661523352688-300x300.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/08\/I09882-Solving-Optimization-Problems-4-1-e1661523352688-150x150.png 150w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/p>\n<p>You may recognize this as a parabola, but the important feature to notice is that it has a maximum point. We can see pretty clearly that this is the highest point on the shape, but it also has a special characteristic which makes it what we call a <a href=\"https:\/\/www.mometrix.com\/academy\/first-derivative-test\/\"class=\"ylist\">critical point<\/a>. <\/p>\n<p>What is the slope of the tangent line at this exact point? Since the line is changing directions from upward to downward, the slope at this exact point is zero, and the tangent line is flat.<\/p>\n<p>This would also be true of a minimum point on a curve. At an arc\u2019s <em>lowest point<\/em>, the line will switch direction from downward to upward and the tangent line will have slope equal to zero. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-138832\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/08\/I09883-Solving-Optimization-problems-5-1-e1661523361698.png\" alt=\"Parabola with an exact point of 0 and a flat tangent line\" width=\"400\" height=\"400\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/08\/I09883-Solving-Optimization-problems-5-1-e1661523361698.png 400w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/08\/I09883-Solving-Optimization-problems-5-1-e1661523361698-300x300.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/08\/I09883-Solving-Optimization-problems-5-1-e1661523361698-150x150.png 150w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/p>\n<p>In other words, at these critical points, the value of a function\u2019s derivative equal 0.<\/p>\n<p>Critical points also exist where a function\u2019s derivative does not exist, such as points where there are breaks in continuity or when a function has \u201csharp\u201d turns.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/critical-point.webp\" alt=\"Two grid graphs with red curves and black arrows showing increasing and decreasing trends, pointing at fluctuations in the curves.\" width=\"637.2\" height=\"209.1\" class=\"aligncenter size-full wp-image-238240\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/critical-point.webp 2124w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/critical-point-300x98.webp 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/critical-point-1024x336.webp 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/critical-point-768x252.webp 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/critical-point-1536x504.webp 1536w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2025\/01\/critical-point-2048x672.webp 2048w\" sizes=\"(max-width: 2124px) 100vw, 2124px\" \/><\/p>\n<p>Because of this, we can find minimum and maximum values of a function within particular domains by searching for points where the derivative of the function equals zero or does not exist.<\/p>\n<p>For example, the function \\(f(x)=-x^{2}+4x\\) is a downward-opening parabola. We can find the maximum of the function by taking the derivative of \\(f\\) and solving for the \\(x\\) value that makes it go to zero or not exist.<\/p>\n<p>Let\u2019s start by taking the derivative of \\(f\\).<\/p>\n<div class=\"examplesentence\">\\(f'(x)=-2x+4\\)<\/div>\n<p>\n&nbsp;<br \/>\nSince there is no point that would make this undefined, we set \\(f&#8217;\\) equal to zero and solve for \\(x\\).<\/p>\n<div class=\"examplesentence\">\\(0=-2x+4\\)<br \/>\n\\(-4=-2x\\)<br \/>\n\\(2=x\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo the maximum for \\(f(x)\\) occurs at \\(x=2\\). We can then evaluate \\(f(2)\\) to find the height.<\/p>\n<div class=\"examplesentence\">\\(f(2)=-(2)^{2}+4(2)\\)\\(=-4+8=4\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo this function has a critical point at \\((2,4)\\).<\/p>\n<p>Let\u2019s return to the example of the rancher now. We know that he needs to use all 400 meters of fencing materials, so the perimeter of the enclosed field must be \\(2l+2w=400\\), where \\(l\\) and \\(w\\) represent length and width respectively.<\/p>\n<p>The area of a rectangle is equal to length times width, but in order to find a critical point, we need to write a function for area in terms of only one variable. Let\u2019s use \\(l\\) as that variable, and we\u2019ll solve for \\(w\\) in terms of \\(l\\) to make the area function.<\/p>\n<div class=\"examplesentence\">\\(2l+2w=400\\)<br \/>\n\\(2w=400-2l\\)<br \/>\n\\(w=200-l\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo the area in terms of \\(l\\) is \\(l \\cdot (200-l)\\).<\/p>\n<div class=\"examplesentence\">\\(A(l)=l\\cdot (200-l)\\).<\/div>\n<p>\n&nbsp;<br \/>\nSimplifying, this equals \\(200l-l^{2}\\).<\/p>\n<p>Remember, in order to find the dimensions that maximize the field\u2019s area, we need to locate a critical point of this function. Let\u2019s take the derivative of \\(A\\).<\/p>\n<div class=\"examplesentence\">\\(A'(l)=200-2l\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe can see that \\(A&#8217;\\) is a linear function. This means it is continuous, and we can solve for where \\(A&#8217;\\) equals zero.<\/p>\n<div class=\"examplesentence\">\\(0=200-2l\\)<br \/>\n\\(-200=-2l\\)<br \/>\n\\(100=l\\)<\/div>\n<p>\n&nbsp;<br \/>\nThere is a critical point when length of the field is equal to 100 meters, which indicates that the area of the field is maximized when the length equals 100 meters and, consequently, when the width equals 100 meters as well. The area of the field with these dimensions is 10,000 square meters.<\/p>\n<h2><span id=\"Optimization_Problem_2\" class=\"m-toc-anchor\"><\/span>Optimization Problem #2<\/h2>\n<p>\nLet\u2019s work through another example. Lisa is working on a geometry project and is making shapes from pliable metal wire. She has 8 feet of wire that she wants to cut into two separate pieces. With one piece she will make a circle, and with the other she will make a square. In order to maximize the total area of the circle and square, how long should the piece of wire for the square be?<\/p>\n<p>Since we are trying to minimize area, we need to come up with a formula for area. This formula needs to be written in terms of only one variable so that we can take its derivative and find a critical point.<\/p>\n<p>Let\u2019s choose the variable to be the letter \\(x\\), representing the length of the wire used for the square. Since the wire is 8 feet long, this restricts the possible values of \\(x\\) to the domain \\(0 \\leq x \\leq 8\\). We know that the total area must include the circle\u2019s area as well as the square\u2019s, so \\(A(x)\\) will look something like this:<\/p>\n<div class=\"examplesentence\">\\(A(x)\\)\\(=\\text{circle&#8217;s area}+\\text{square&#8217;s area}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe area of a square is equal to its side length squared, and the side length of Lisa\u2019s square will equal one-fourth of \\(x\\). So the square\u2019s area will equal \\((\\frac{x}{4})^2\\), or \\(\\frac{x^2}{16}\\). <\/p>\n<p>To get the area of the circle, we need to remember that the length of leftover wire from the square will equal \\(8-x\\) feet. This will make up the circumference of Lisa\u2019s circle. The circumference of a circle equals \\(2\\pi r\\), and the area of a circle equals \\(\\pi r^{2}\\).<\/p>\n<p>Let\u2019s get the value of the radius, \\(r\\), by dividing \\(8-x\\) by \\(2\\pi\\).<\/p>\n<div class=\"examplesentence\">\\(r=\\frac{8-x}{2\\pi }\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, we can square this and multiply by \\(\u03c0\\) to get the circle\u2019s area.<\/p>\n<div class=\"examplesentence\">\\(circle&#8217;s area=\\frac{(8-x)^{2}}{4\\pi }\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we have the formulas for the circle\u2019s area and the square\u2019s area individually. Let\u2019s put them together to get the area function.<\/p>\n<div class=\"examplesentence\">\\(A(x)=\\frac{(8-x)^2}{4\\pi}+\\frac{x^2}{16}\\)<\/div>\n<p>\n&nbsp;<\/p>\n<p>Let&#8217;s expand the numerator of the first term.<\/p>\n<div class=\"examplesentence\">\\(A(x)=\\frac{64-16x+x^{2}}{4\\pi }+\\frac{x^{2}}{16}\\)<\/div>\n<p>\n&nbsp;<br \/>\nTo find the derivative of the first term, we could use the <a href=\"https:\/\/www.mometrix.com\/academy\/product-and-quotient-rule\/\" class=\"ylist\">quotient rule<\/a>. However, since the bottom is a constant, we can just factor it out and take the derivative of the top:<\/p>\n<div class=\"examplesentence\">\\(A'(x)=\\frac{1}{4\\pi }(-16+2x)\\)\u2026<\/div>\n<p>\n&nbsp;<\/p>\n<p>Then we can add the derivative of the second term, which is \\(\\frac{x}{8}\\).<\/p>\n<div class=\"examplesentence\">\\(A'(x)=\\frac{1}{4\\pi } (-16+2x)+\\frac{x}{8}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis resulting derivative is a linear function, and we can solve for where it equals 0.<\/p>\n<div class=\"examplesentence\">\\(0=\\frac{1}{4\\pi }(-16+2x)+\\frac{x}{8}\\)<br \/>\n\\(0=-\\frac{4}{\\pi }+\\frac{x}{2\\pi }+\\frac{x}{8}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIsolate the \\(x\\) terms.<\/p>\n<div class=\"examplesentence\">\\(\\frac{4}{\\pi}=\\frac{x}{2\\pi }+\\frac{x}{8}\\)<\/div>\n<p>\n&nbsp;<br \/>\nEstablish a common denominator of \\(8\\pi\\).<\/p>\n<div class=\"examplesentence\">\\(\\frac{4}{\\pi}=\\frac{4x}{8\u03c0}+\\frac{\u03c0x}{8\u03c0}\\)<br \/>\n\\(\\frac{4}{\u03c0}=\\frac{4x+\u03c0x}{8\u03c0}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSimplify to find the exact value of \\(\\frac{32}{4+\\pi}\\), which is approximately 4.5.<\/p>\n<div class=\"examplesentence\">\\(32=4x+\u03c0x\\)<br \/>\n\\(32=x(4+\u03c0)\\)<br \/>\n\\(\\frac{32}{4+\u03c0}=x\\)<br \/>\n\\(x\\approx 4.5\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis indicates that Lisa would cut roughly 4.5 feet for the square, leaving around 3.5 feet for the circle.<\/p>\n<p>Remember, even though we have found this critical point, we also must consider the endpoints of the domain. In this case, we should check \\(A(0)\\) and \\(A(8)\\).<\/p>\n<p>\\(A(0)\\) represents the total area if Lisa uses all the wire for the circle, and this area is approximately 5.093 square feet.<\/p>\n<div class=\"examplesentence\">\\(A(0)=\\frac{(8-0)^{2}}{4\\pi }+\\frac{1}{16}(0)^{2}=\\frac{16}{\\pi }\\)\\(\\approx 5.093\\text{ square feet}\\)<\/div>\n<p>\n&nbsp;<br \/>\n\\(A(8)\\) represents the total area if Lisa uses all the wire for the square, and this area is 4 square feet.<\/p>\n<div class=\"examplesentence\">\\(A(8)=\\frac{(8-8)^{2}}{4\\pi }+\\frac{1}{16}(8)^{2}\\)\\(=4\\text{ square feet}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf Lisa cuts the wire at the critical point, she will have a total area of approximately 2.24 square feet.<\/p>\n<div class=\"examplesentence\">\\(A(\\frac{32}{4+\\pi })=\\frac{(8-\\frac{32}{4+\\pi })^{2}}{4\\pi }+\\frac{1}{16}(\\frac{32}{4+\\pi })^{2}\\)\\(\\approx 2.24\\text{ square feet}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIt seems that Lisa will have the smallest area if she uses the critical point and dedicates about 4.5 feet of wire to the square!<\/p>\n<hr>\n<p>\nWhenever you see the words \u201cmaximize\u201d or \u201cminimize\u201d in a calculus problem, these are signals that indicate an optimization problem is at hand. The first step in solving these is writing a formula for what is being optimized in terms of only one variable. Then, take the derivative, look for critical points, and test these against the endpoints of the domain to find the true maximum or minimum.<\/p>\n<p>This may seem like a daunting process at first, but with some practice you\u2019ll get the hang of it.<\/p>\n<p>That&#8217;s all for this review. Thanks for watching, and happy studying!<\/p>\n<\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/calculus\/\">Return to Calculus Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Calculus Videos<\/p>\n","protected":false},"author":22,"featured_media":237610,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-131848","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-calculus-videos","7":"page_category-video-pages-for-study-course-sidebar-ad","8":"page_type-video","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/131848","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=131848"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/131848\/revisions"}],"predecessor-version":[{"id":281342,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/131848\/revisions\/281342"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/237610"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=131848"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}