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Equations - Homogeneous</a></label></li></ul></div> <div class="hidden-xs hidden-sm"> <button class="btn btn-default btn-block text-xs-center" data-toggle="modal" data-target="#modal-feedback" style="margin-bottom: 10px"><i class="fa fa-send"></i> Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> Optimisation <a href="#" class="mark-page-favorite pull-right" data-pid="695" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../mathsanalysis.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../550/calculus.html">Calculus</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Optimisation</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 30 minutes"><i class="fa fa-clock-o"></i> 30'</span> </ol> <article id="main-article"> <p><img alt="" src="../../files/differentiation/optimisation/main-1.png" style="float: left; width: 100px; height: 102px;">Optimisation questions involve finding the best solution to a problem. Usually, that is the maximum or minimum value of a function. Questions on this topic require you to take information from a practical problem and write this as a function, then use differentiation to find the maximum or minimum value of this function by solving where the gradient is equal to zero. Often questions on this topic seem to have two variables, but you need to take a piece of information that fixes one of the variables and substitute this to create a function with one variable (although HL students might have to use <a href="../741/implicit-differentiation.html" title="Implicit Differentiation">Implicit Differentiation</a>!)</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <p>On this page, you should learn about</p> <ul> <li>solving optimisation problems using differentiation</li> </ul> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <p>The following videos will help you understand all the concepts from this page</p> <div class="panel panel-yellow panel-has-colored-body panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Optimum Can</p> </div> </div> <div class="panel-body"> <div class="smart-object center" data-id="494"> <h3>Optimum Can</h3> <p>In the following video we look at an example of using differentiation to find the best possible can for a drinks manufacturer.</p> <p><em><strong>A drinks manufacturer wants a design for a new can for their mini drinks collection, volume=170ml.Your job is to find the can that will contain a volume of 170ml = 170cm3 ≈54πcm3 and that will minimize the amount of aluminium used.</strong></em></p> <p>It might help you to visualize the can by playing with the following applet</p> <p style="text-align: center"><iframe height="580px" scrolling="no" src="https://www.geogebra.org/material/iframe/id/e78gybtw/width/529/height/580/border/888888/sfsb/true/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false" style="border:0px;" title="Volume of Cone2" width="529px"></iframe></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/293605550"></iframe></div> <h4><span></span><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span><span></span> Notes from the video</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><span class="fa fa-print" style="color:rgb(0, 0, 0);font-size:14px;"></span> Print from <a href="../../files/differentiation/optimisation/optimisation_cylinder.pdf" target="_blank">here</a></p> <p style="text-align: center;"><iframe align="middle" frameborder="0" height="480" scrolling="yes" src="../../files/differentiation/optimisation/optimisation_cylinder.pdf" width="640"></iframe></p> </section> </div> </div> </div> <div class="panel panel-yellow panel-has-colored-body panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Optimisation - Rectangle in a Graph</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="493"> <p>The following is an example of an optimisation problem</p> <p>A rectangle ABCD is drawn so that its lower vertices are on the x-axis and its upper vertices are on the curve <span class="math-tex">\(f(x)=\sqrt{6-3x^2}\)</span> as shown in the following diagram</p> <p><img alt="" src="../../files/differentiation/optimisation/image-recatngle.png" style="width: 332px; height: 285px;"></p> <p>Let OA = x<br> The area of this rectangle is denoted by A.</p> <p style="margin-left:.375in;">(a) Write down an expression for A in terms of x.</p> <p style="margin-left:.375in;">(b) Find the maximum value of A.</p> <hr class="hidden-separator"> <p>You may find the following applet useful to get an understanding of the problem.</p> <p style="text-align: center"><iframe height="314px" scrolling="no" src="https://www.geogebra.org/material/iframe/id/tmxwe9as/width/490/height/314/border/888888/sfsb/true/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false" style="border:0px;" title="Rectangle in a graph" width="490px"></iframe></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/299933703"></iframe></div> <h4><span></span><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span><span></span> Notes from the video</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><span class="fa fa-print" style="color:rgb(0, 0, 0);font-size:14px;"></span> Print from <a href="../../files/differentiation/optimisation/rectangle-in-a-graph.pdf" target="_blank">here</a></p> <p style="text-align: center;"><iframe align="middle" frameborder="0" height="480" scrolling="yes" src="../../files/differentiation/optimisation/rectangle-in-a-graph.pdf" width="640"></iframe></p> </section> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> <p> </p> <p> </p> <p> </p> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander pull-right" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <p>Here is a quiz that practises setting up the problem before you use Calculus to solve. You are strongly recommended to have something to write with whilst doing these questions. You may want to draw sketches of shapes and graphs to help you think through the problems. You are not required to do any Calculus in this quiz!</p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#4b5d5bfb"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="4b5d5bfb"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Optimisation <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="11-566-695" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>120m of fencing is used to make an enclosure against a wall where <em><strong>x</strong></em> represents the width of the enclosure.</p><p>What is the fomula for the area? <b><i>A </i></b>of the pen</p><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q1.1.png" style="width: 400px; height: 203px;"></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> A = x(60 - x)</label></p><p><label class="radio"><input class="c" type="radio"> A = x(120 - 2x)</label></p><p><label class="radio"><input type="radio"> A = 120 - x</label></p><p><label class="radio"><input type="radio"> A = x(120 - x)</label></p></div><div class="q-explanation"><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q1.png" style="width: 350px; height: 166px;"></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A rectangle has area 20cm². If one side of the rectangle is <strong><em>x</em></strong>, find a formula for the perimeter of the rectangles</p></div><div class="q-answer"><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\(2x + \frac{40}{x}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(2x + \frac{20}{x}\)</span></label></p><p><label class="radio"><input type="radio"> x(20 - x)</label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(x + \frac{20}{x}\)</span></label></p></div><div class="q-explanation"><p><img alt="" height="176" src="../../files/differentiation/optimisation/quiz1/q10-becomes-q2.png" width="326"></p><p>Perimeter = <span class="math-tex">\(x + \frac{20}{x}+x + \frac{20}{x}\)</span></p><p>P = <span class="math-tex">\(2x + \frac{40}{x}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>An open box is made from a piece of paper 20cm by 20cm by removing a square from each corner and folding.</p><p>If <strong><em>x</em></strong> represents the side of the square, what is the formula for the volume, <strong><em>V</em></strong> of the box?</p><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q2.png" style="width: 329px; height: 215px;"></p></div><div class="q-answer"><p><label class="radio"><input class="c" type="radio"> V = x(20 - 2x)²</label></p><p><label class="radio"><input type="radio"> V = 20x(20 - x)</label></p><p><label class="radio"><input type="radio"> V = x²(20 - 2x)</label></p><p><label class="radio"><input type="radio"> V = x²(20 - x)</label></p></div><div class="q-explanation"><p>Length = 20 - 2x</p><p>Width = 20 - 2x</p><p>Height = x</p><p>Volume = length x width x height</p><p>Volume = (20 - 2x)(20 - 2x)x</p><p>Volume = x(20 - 2x)²</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>An open box is made from a piece of paper 30cm by 20cm by removing a square from each corner and folding.</p><p>If <strong><em>x</em></strong> represents the side of the square, what is the formula for the volume, <strong><em>V</em></strong> of the box?</p><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q3" style="width: 350px; height: 160px;"></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> V = x(30 - x)(20 - x)</label></p><p><label class="radio"><input class="c" type="radio"> V = x(30 - 2x)(20 - 2x)</label></p><p><label class="radio"><input type="radio"> V = x²(30 - 2x)(20 - 2x)</label></p><p><label class="radio"><input type="radio"> You cannot make this box from a rectangular piece of paper</label></p></div><div class="q-explanation"><p>Length = 30 - 2x</p><p>Width = 20 - 2x</p><p>Height = x</p><p>Volume = length x width x height</p><p>Volume = (30 - 2x)(20 - 2x)x</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A piece of wire 50cm long is bent into the shape of a rectangle. If <strong><em>x</em></strong> represents the length of the side of the rectangle, what is the formula for the area, <strong><em>A</em></strong> of the rectangle created?</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> A = x(50 - 2x)</label></p><p><label class="radio"><input type="radio"> A = x(50 - x)</label></p><p><label class="radio"><input class="c" type="radio"> A = x(25 - x)</label></p><p><label class="radio"><input type="radio"> A = 50x</label></p></div><div class="q-explanation"><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q4.png" style="width: 400px; height: 159px;"></p><p>Width = 25-x</p><p>Area = x(25 - x)</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A closed cuboid box has has volume 1000cm<sup>3</sup></p><p>If the base is square with side <strong><em>x</em></strong> cm, find the <strong>total surface area</strong> of the cuboid</p><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q2.png" style="width: 329px; height: 215px;"></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(4x^2+\frac{2000}{x}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(2x^2+\frac{4000}{x^2}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(x^2+\frac{4000}{x}\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\(2x^2+\frac{4000}{x}\)</span></label></p></div><div class="q-explanation"><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q5a.png" style="width: 300px; height: 200px;"></p><p>Volume = x²h</p><p>1000 = x²h</p><p><span class="math-tex">\(h=\frac{1000}{x^2}\)</span></p><p>Total surface area <span class="math-tex">\(=2\times x^2+4 \times x\times \frac{1000}{x^2}\)</span></p><p>Total surface area <span class="math-tex">\(=2x^2+\frac{4000}{x}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q6.png" style="width: 350px; height: 238px;"></p><p>ABC is a right-angled triangle. The sum of the lengths AB and AC is 10cm.</p><p>Find the area of the triangle if AB = <strong><em>x</em></strong></p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(x(10-x)\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\(\frac{x}{2}\sqrt{100-20x}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(\frac{x}{2}({10-x})\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(\frac{x}{2}\sqrt{100-2x²}\)</span></label></p></div><div class="q-explanation"><p>We can use Pythagoras' Theorem to find the missing length</p><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q6a.png" style="width: 550px; height: 160px;"></p><p>Area =<span class="math-tex">\(\frac{x}{2}\sqrt{100-20x}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A rectangle is drawn so that its lower vertices lie on the x axis and its upper vertices lie on the curve</p><p>y = a² - x²</p><p>Find an expression for the area of the rectangle.</p></div><div class="q-answer"><p><label class="radio"><input class="c" type="radio"> 2x(a² - x²)</label></p><p><label class="radio"><input type="radio"> a²x</label></p><p><label class="radio"><input type="radio"> x(a² - x²)</label></p><p><label class="radio"><input type="radio"> 2a²x</label></p></div><div class="q-explanation"><p>It is a good idea to draw a sketch of the graph.</p><p>The width of the rectangle is 2x</p><p>The height of the rectangle is the value of the function a² - x²</p><p>Therefore, the area = 2x(a² -x²)</p><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q7a.png" style="width: 350px; height: 263px;"></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A rectangle is drawn so that its lower vertices lie on the x axis and its upper vertices lie on the curve</p><p>y = f(x)</p><p><span class="math-tex">\(f(x)=\left\{\begin{matrix}e^x\ \ \ \ \ \ \ \ x<0\\e^{-x}\ \ \ \ \ \ \ \ x\geq0\\\end{matrix}\right.\)</span></p><p>Find an expression for the area of the rectangle.</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(xe^{-x}\)</span></label><label class="radio"></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(2xe^{x}\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\(2xe^{-x}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(xe^{x}\)</span></label></p></div><div class="q-explanation"><p>It is a good idea to draw a sketch of the graph.</p><p>The width of the rectangle is 2x</p><p>The height of the rectangle is the value of the function <span class="math-tex">\(e^{-x}\)</span></p><p>Therefore, the area = <span class="math-tex">\(2xe^{-x}\)</span></p><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q8.png" style="width: 350px; height: 218px;"></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q9.png" style="width: 400px; height: 339px;"></p><p>A half cylinder has volume = 400</p><p>Find a formula for the <strong><em>A</em></strong>, surface area of the solid in terms of <strong><em>r</em></strong>, the radius of the corss section</p></div><div class="q-answer"><p><label class="radio"><input type="radio"> <span class="math-tex">\(A=\pi r^2+\frac{400}{ r}\)</span></label></p><p><label class="radio"><input class="c" type="radio"> <span class="math-tex">\(A=\pi r^2+\frac{1600}{\pi r}+\frac{800}{ r}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(A=\pi r^2+\frac{800}{\pi r}+\frac{400}{ r}\)</span></label></p><p><label class="radio"><input type="radio"> <span class="math-tex">\(A=\pi r^2+\frac{800}{ r}\)</span></label></p></div><div class="q-explanation"><p>Let the length of the cylinder = <strong><em>l</em></strong></p><p><img alt="" src="../../files/differentiation/optimisation/quiz1/q9a.png" style="width: 350px; height: 283px;"></p><p>Volume = <span class="math-tex">\(\frac{\pi r^2}{2}l\)</span></p><p><span class="math-tex">\(\frac{\pi r^2}{2}l=400\)</span></p><p><span class="math-tex">\(l=\frac{800}{\pi r^2}\)</span></p><p>The surface area of the solid is made up of</p><p>2 semi-circular ends = <span class="math-tex">\(2\times\frac{\pi r^2}{2}\)</span></p><p>+</p><p>rectangular base = <span class="math-tex">\(2rl\)</span></p><p>+</p><p>curved suface = <span class="math-tex">\(\pi rl\)</span></p><p>Total surface area = <span class="math-tex">\(\pi r^2+2rl+\pi rl\)</span></p><p><span class="math-tex">\(A=\pi r^2+2r\frac{800}{\pi r^2}+\pi r\frac{800}{\pi r^2}\)</span></p><p><span class="math-tex">\(A=\pi r^2+\frac{1600}{\pi r}+\frac{800}{ r}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i> Prev </button> </div> <div class="pull-right pull-xs-none"> <button class="btn btn-success btn-xs-block text-xs-center btn-results" style="display: none"> <i class="fa fa-bar-chart"></i> Check Results </button> <button class="btn btn-default d-xs-none btn-next"> Next <i class="fa fa-arrow-right"></i> </button> <button class="btn btn-default btn-xs-block text-xs-center btn-close" data-dismiss="modal" style="display: none"> Close </button> </div> </div> </div> </div> </div></div> </div> <div class="panel-footer"> <div> <p>text</p> </div> </div> </div> <div class="panel panel-has-colored-body panel-default"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Exam-style Questions</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-default panel-has-border"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 1</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="495"> <p><img class="sibico" src="../../../img/sibico/hl-orange.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL moderate"> <img class="sibico" src="../../../img/sibico/sl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="SL difficult"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p><img alt="" src="../../files/differentiation/optimisation/esq1.jpg" style="float: left; width: 640px; height: 254px;"></p> <hr class="hidden-separator"> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden">Area of a cylinder , <span class="math-tex">\(A=2\pi r^2+2\pi rh\)</span><content></content></section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p><span class="fa fa-print" style="color:rgb(0, 0, 0);font-size:14px;"></span> Print from <a href="../../files/differentiation/optimisation/esq_optimisation1.pdf" target="_blank">here</a></p> <p><iframe align="middle" frameborder="0" height="480" scrolling="yes" src="../../files/differentiation/optimisation/esq_optimisation1.pdf" width="640"></iframe></p> </section> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 2</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="889"> <p><img class="sibico" src="../../../img/sibico/hl-orange.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL moderate"> <img class="sibico" src="../../../img/sibico/sl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="SL difficult"> <img class="sibico" src="../../../img/sibico/no-calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="No calculator"></p> <p>A container is made from a cylinder and a hemisphere. The radius of the cylinder is <strong><em>r</em></strong> m and the height is <strong><em>h</em></strong>. The volume of the container is <span class="math-tex">\(45\pi\)</span></p> <p><img alt="" src="../../files/differentiation/optimisation/esq2.jpg" style="width: 261px; height: 329px;"></p> <p>a) Find an expression for the height of the cylinder in terms of <strong><em>r</em></strong></p> <p>b) Show that the surface area of the container, <span class="math-tex">\(A=\frac{5 \pi r^2}{3}+\frac{90 \pi}{r}\)</span></p> <p>c) Hence, find the values of <strong><em>r</em></strong> and <strong><em>h</em></strong> that give the minimum surface area of the container</p> <p>* Volume of a sphere = <span class="math-tex">\(\frac{4}{3}\pi r^3\)</span></p> <p>** Surface area of a sphere = <span class="math-tex">\(4\pi r^2\)</span></p> <hr class="hidden-separator"> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>The volume of a hemisphere = <span class="math-tex">\(\frac{2}{3}\pi r^3\)</span><content></content></p> <p>The curved surface area of a hemisphere = <span class="math-tex">\(2\pi r^2\)</span></p> <p>The surface area of the container = <span class="math-tex">\(2\pi r^2+\pi r^2+2\pi rh\)</span></p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>a) The volume of a hemisphere = <span class="math-tex">\(\frac{2}{3}\pi r^3\)</span></p> <p>The volume of the cylinder = <span class="math-tex">\(\pi r^2h\)</span></p> <p>Hence, the volume of the container = <span class="math-tex">\(\frac{2}{3}\pi r^3+\pi r^2h\)</span></p> <p><span class="math-tex">\(\frac{2}{3}\pi r^3+\pi r^2h=45\pi\)</span></p> <p>We can divide through by <span class="math-tex">\(\pi\)</span> and rearrange to make <strong><em>h</em></strong> the subject of the formula</p> <p><span class="math-tex">\(\frac{2}{3} r^3+ r^2h=45\)</span></p> <p><span class="math-tex">\(r^2h=45-\frac{2}{3} r^3\)</span></p> <p><span class="math-tex">\(h=\frac{45}{r^2}-\frac{2}{3} r\)</span></p> <p>b) The surface area of the container = <span class="math-tex">\(2\pi r^2+\pi r^2+2\pi rh\)</span></p> <p><span class="math-tex">\(A=3\pi r^2+2\pi rh\)</span></p> <p><span class="math-tex">\(A=3\pi r^2+2\pi r(\frac{45}{r^2}-\frac{2}{3} r)\)</span></p> <p><span class="math-tex">\(A=\frac{9\pi r^2}{3}-\frac{4 \pi r^2}{3}+\frac{90 \pi}{r}\)</span></p> <p><span class="math-tex">\(A=\frac{5\pi r^2}{3}+\frac{90 \pi}{r}\)</span></p> <p>c) <span class="math-tex">\(\frac{dA}{dr}=\frac{10\pi r}{3}-\frac{90 \pi}{r^2}\)</span></p> <p>Minimimum occurs when <span class="math-tex">\(\frac{dA}{dr}=0\)</span></p> <p><span class="math-tex">\(\frac{10\pi r}{3}-\frac{90 \pi}{r^2}=0\)</span></p> <p><span class="math-tex">\(\frac{10\pi r}{3}=\frac{90 \pi}{r^2}\)</span></p> <p><span class="math-tex">\(r^3=27\)</span></p> <p>r = 3</p> <p><span class="math-tex">\(h=\frac{45}{3^2}-\frac{2}{3} \times 3\)</span></p> <p>h = 5 - 2</p> <p>h = 3</p> <p>We can verify that this gives the minimum value by finding the 2nd derivative</p> <p><span class="math-tex">\(\frac{dA}{dr}=\frac{10\pi r}{3}-\frac{90 \pi}{r^2}\)</span></p> <p><span class="math-tex">\(\frac{d^2A}{dr^2}=\frac{180 \pi}{r^3}\)</span></p> <p>When r = 3, <span class="math-tex">\(\frac{d^2A}{dr^2}>0\)</span> , hence <em><strong>A</strong></em> is a minimum</p> </section> <h4> </h4> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-default panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Question 3</p> </div> </div> <div class="panel-body"> <div> <div class="smart-object center" data-id="888"> <p><img class="sibico" src="../../../img/sibico/hl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="HL difficult"> <img class="sibico" src="../../../img/sibico/sl-red.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="SL difficult"> <img class="sibico" src="../../../img/sibico/calc.svg" style="height:1.25em;width: auto;vertical-align:text-bottom" title="Calculator"></p> <p>The diagram below shows the graph of the functions f(x) = sinx and g(x) = 2sinx</p> <p><img alt="" src="../../files/differentiation/optimisation/esq3_1.jpg" style="width: 500px; height: 376px;"></p> <p>A rectangle ABCD is placed in between the two functions as shown so that B and C lie on <strong><em>g</em></strong> , BC is parallel to the x axis and the local minima of the function <em>f</em> lies on AD.</p> <p>Let NA = <strong><em>x</em></strong></p> <p>a) Find an expression for the height of the rectangle AB</p> <p>b) Show that the area of the rectangle, <strong><em>A</em></strong> can be given by A = 4xcosx - 2x</p> <p>c) Find <span class="math-tex">\(\frac{dA}{dx}\)</span></p> <p>d) Find the maximum value of the area of the rectangle.</p> <h4><span class="fa fa-support" style="color:rgb(0, 0, 0);"></span> Hint</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>a) The y coordinate of B is <span class="math-tex">\(g(\frac{\pi}{2}+x)=2sin(\frac{\pi}{2}+x)\)</span></p> <p>b) Note that <span class="math-tex">\(sin(x+\frac{\pi}{2})=cosx\)</span><content></content></p> </section> <h4><span class="fa fa-pencil" style="color:rgb(0, 0, 0);"></span> Full Solution</h4> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <p>a) The y cordinate of A is 1 (the maximum value of the function f(x) = sinx)</p> <p>The y coordinate of B is <span class="math-tex">\(g(\frac{\pi}{2}+x)=2sin(\frac{\pi}{2}+x)\)</span></p> <p>The height of the rectangle, AB = <span class="math-tex">\(2sin(\frac{\pi}{2}+x)-1\)</span></p> <p>Note that <span class="math-tex">\(sin(x+\frac{\pi}{2})=cosx\)</span></p> <p>Therfore, AB = 2cosx - 1</p> <p><img alt="" src="../../files/differentiation/optimisation/esq3a.jpg" style="width: 400px; height: 248px;"></p> <p>b) The width of the rectangle = 2x</p> <p>Hence, the area of the rectangle, A = 2x(2cosx - 1) = 4xcosx - 2x</p> <p>c) To find <span class="math-tex">\(\frac{dA}{dx}\)</span>, we need to use the <a href="../743/product-and-quotient-rule.html">Product Rule</a></p> <p><span class="math-tex">\(\frac{dA}{dx}=4cosx - 4xsinx - 2\)</span></p> <p>d) The maximum value of the area of the rectangle occurs where <span class="math-tex">\(\frac{dA}{dx}=0\)</span></p> <p>We can use our graphical calculator to solve this. The value occurs when x <span class="math-tex">\(\approx\)</span> 0.592</p> <p>A<sub>max</sub> <span class="math-tex">\(\approx\)</span> 0.781</p> </section> <h4> </h4> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="page-container panel-self-assessment" data-id="695"> <div class="panel-heading">MY PROGRESS</div> <div class="panel-body understanding-rate"> <div class="msg"></div> <label class="label-lg">Self-assessment</label><p>How much of <strong>Optimisation</strong> have you understood?</p><div class="slider-container text-center"><div id="self-assessment-slider" class="sib-slider self-assessment " data-value="1" data-percentage=""></div></div> <label class="label-lg">My notes</label> <textarea name="page-notes" class="form-control" rows="3" placeholder="Write your notes here..."></textarea> </div> <div class="panel-footer text-xs-center"> <span id="last-edited" 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