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href="../1078/calculus.html">Calculus</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">5.6 & 5.7 Stationary Points and 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"> <h2><img alt="" src="../../files/CLS/spo/spo-1.png" style="float: left; width: 100px; height: 100px;">Max, Min and Optimisation</h2> <p>This is the part where we learn about how we use calculus to solve different kinds of problems. So far we have spent a lot of time learning about how calculus helps us to work out the gradient of given function for any given point. In this section we focus on the points where the gradient is zero. These are called the local maxima and minima and crucial in what we call 'optimisation' where we try to work out how to maximise profits or minimise costs and so on.</p> <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><span id="docs-internal-guid-495bdd37-7fff-8f70-eda7-eb9277ce534c"></span></p> <p dir="ltr"><span id="docs-internal-guid-495bdd37-7fff-8f70-eda7-eb9277ce534c">In this unit you should learn to…</span></p> <ul style="list-style-type:disc;"> <li aria-level="1" dir="ltr"> <p dir="ltr" role="presentation"><span id="docs-internal-guid-495bdd37-7fff-8f70-eda7-eb9277ce534c">Find where the gradient of a function is zero</span></p> </li> <li aria-level="1" dir="ltr"> <p dir="ltr" role="presentation">Find local maximum and minimum points</p> </li> <li aria-level="1" dir="ltr"> <p dir="ltr" role="presentation">Use calculus to solve an optimisation problem</p> </li> </ul> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <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"> <div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Slides Gallery</p> </div> </div> <div class="panel-body"> <div> <p>Use these slides to review the material and key points covered in the videos.</p> <div id="carousel-265" class="dynamic-gallery carousel slide" data-id="265"><div class="carousel-inner" role="listbox"><div class="item active"><a class="fancy" href="../../../std-galleries/15-265/opt-nologos001.jpeg" data-fancybox="gallery-265" title="" data-caption=""><img alt="" src="../../../std-galleries/15-265/opt-nologos001.jpeg"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/15-265/opt-nologos002.jpeg" data-fancybox="gallery-265" title="" data-caption=""><img alt="" src="../../../std-galleries/15-265/opt-nologos002.jpeg"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/15-265/opt-nologos003.jpeg" data-fancybox="gallery-265" title="" data-caption=""><img alt="" src="../../../std-galleries/15-265/opt-nologos003.jpeg"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/15-265/opt-nologos004.jpeg" data-fancybox="gallery-265" title="" data-caption=""><img alt="" src="../../../std-galleries/15-265/opt-nologos004.jpeg"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/15-265/opt-nologos005.jpeg" data-fancybox="gallery-265" title="" data-caption=""><img alt="" src="../../../std-galleries/15-265/opt-nologos005.jpeg"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/15-265/opt-nologos006.jpeg" data-fancybox="gallery-265" title="" data-caption=""><img alt="" src="../../../std-galleries/15-265/opt-nologos006.jpeg"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/15-265/opt-nologos007.jpeg" data-fancybox="gallery-265" title="" data-caption=""><img alt="" src="../../../std-galleries/15-265/opt-nologos007.jpeg"></a></div><div class="item "><a class="fancy" href="../../../std-galleries/15-265/opt-nologos008.jpeg" data-fancybox="gallery-265" title="" data-caption=""><img alt="" src="../../../std-galleries/15-265/opt-nologos008.jpeg"></a></div></div><a class="left carousel-control" href="#carousel-265" role="button" data-slide="prev"><i class="fa fa-fw fa-chevron-left"></i></a><a class="right carousel-control" href="#carousel-265" role="button" data-slide="next"><i class="fa fa-fw fa-chevron-right"></i></a></div><ol class="std-carousel-indicators"><li data-index="0"><img title="Click to view" src="../../../std-galleries/15-265/opt-nologos001-thumb128.jpg"><li><li data-index="1"><img title="Click to view" src="../../../std-galleries/15-265/opt-nologos002-thumb128.jpg"><li><li data-index="2"><img title="Click to view" src="../../../std-galleries/15-265/opt-nologos003-thumb128.jpg"><li><li data-index="3"><img title="Click to view" src="../../../std-galleries/15-265/opt-nologos004-thumb128.jpg"><li><li data-index="4"><img title="Click to view" src="../../../std-galleries/15-265/opt-nologos005-thumb128.jpg"><li><li data-index="5"><img title="Click to view" src="../../../std-galleries/15-265/opt-nologos006-thumb128.jpg"><li><li data-index="6"><img title="Click to view" src="../../../std-galleries/15-265/opt-nologos007-thumb128.jpg"><li><li data-index="7"><img title="Click to view" src="../../../std-galleries/15-265/opt-nologos008-thumb128.jpg"><li></li></ol> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>1. Gradient zero</p> </div> </div> <div class="panel-body"> <div> <p>This video is about finding the value of x when the function has a gradient of zero and how we use calculus to do this.</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/595235518"></iframe></div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>2. Finding the points where the gradient is zero.</p> </div> </div> <div class="panel-body"> <div> <p>In this video we extend the idea to functions that have more than one point where the gradient is zero and how to distinguish between them. </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/595236815"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>3. Finding local maximum and minimum points</p> </div> </div> <div class="panel-body"> <div> <p>This video is about finding local maxima and minima and distinguishing between them using the second derivative.</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/595237799"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>4. Optimisation part 1</p> </div> </div> <div class="panel-body"> <div> <p>In this video we get a first look at how an optimisation problem that we need calculus to solve.</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/595238630"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>5. Making models to optimise</p> </div> </div> <div class="panel-body"> <div> <p>Often an optimisation problem requires you to put some information together to first make a model. This video offers some examples and help with that idea. </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/595239303"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>6. Making more complicated models</p> </div> </div> <div class="panel-body"> <div> <p>This is an extension of the above idea, but looks at some of the more complicated examples.</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/595239838"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>7. Optimisation part 2</p> </div> </div> <div class="panel-body"> <div> <p>Now we put these ideas together all in one problem. This video covers making the model.</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/595240633"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>8. Optimisation part 3</p> </div> </div> <div class="panel-body"> <div> <p>Here is the second part of this problem. This part covers the calculus!</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/595242128"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> </div> </div> </div> <div class="panel panel-has-colored-body panel-violet"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Summary</p> </div> </div> <div class="panel-body"> <p><em><strong>Revision Cards</strong></em> - Review this condensed 'key point' revision card to help you check and keep ideas fresh in your mind.</p> <p style="text-align: center;"><img alt="" src="../../files/CLS/spo/c3rc1.001.jpeg" style="width: 700px; height: 525px;"></p> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <div> <h4>Self Checking Quiz</h4> <p>Practice your understanding on these questions. Check your answers when you are done and read the hints where you got stuck. If you find there are still some gaps in your understanding then go back to the videos and slides above. Depending on how confident you are with these ideas, this might take 2-3 hours to complete and you should work with a pen and paper to hand.</p> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <div class="tib-quiz" data-stats="15-330-2402"><div class="label label-default q-number">1</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 1</p><p>Consider the function <span class="math-tex">\(f(x)={ x }^{ 2 }-4x+7\)</span></p><p>then <span class="math-tex">\(f'(x)\quad =\quad ax\quad +\quad b\)</span></p><p>a) What are the values of a and b?</p><p>b) What is the value of x when the gradient is zero.</p><p>c) And so if the coordinates of the vertex are (x,y), what are the values of x and y?</p></div><div class="q-answer"><p>a) a = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> , b = <input type="text" style="height: auto;" data-c="-4"> <span class="review"></span> </p><p>b) x = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> </p><p>c) x = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> , y = <input type="text" style="height: auto;" data-c="3"> <span class="review"></span> </p></div><div class="q-explanation"><p>a) Function differentiates to <span class="math-tex">\(f'(x)\quad =\quad 2x-4\)</span> (careful, since it is -4, but the equation says +b, then b must be negative.</p><p>b) <span class="math-tex">\(when\quad gradient\quad =\quad 0,\\ f'(x)\quad =\quad 2x-4\quad =\quad 0\\ so,\quad 2x\quad =\quad 4\\ and\quad x\quad =\quad 2\)</span></p><p>c) <span class="math-tex">\(At\quad vertex,\quad f'(x)\quad =\quad 0\quad so\quad x\quad =\quad 2\\ Substitute\quad x\quad =\quad 2\quad into\quad f(x)={ x }^{ 2 }-4x+7\\ So,\quad y\quad =\quad 3\)</span> (You might use your table function here)</p></div><div class="actions"><span class="score" data-score="0"></span><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="label label-default q-number">2</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 2</p><p>Consider the function <span class="math-tex">\(f(x)={ x }^{ 2 }+6x+3\)</span></p><p>then <span class="math-tex">\(f'(x)\quad =\quad ax\quad +\quad b\)</span></p><p>a) What are the values of a and b?</p><p>b) What is the value of x when the gradient is zero.</p><p>c) And so if the coordinates of the vertex are (x,y), what are the values of x and y?</p></div><div class="q-answer"><p>a) a = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> , b = <input type="text" style="height: auto;" data-c="6"> <span class="review"></span> </p><p>b) x = <input type="text" style="height: auto;" data-c="-3"> <span class="review"></span> </p><p>c) x = <input type="text" style="height: auto;" data-c="-3"> <span class="review"></span> , y = <input type="text" style="height: auto;" data-c="-6"> <span class="review"></span> </p></div><div class="q-explanation"><p>a) Function differentiates to <span class="math-tex">\(f'(x)\quad =\quad 2x+6\)</span> </p><p>b) <span class="math-tex">\(when\quad gradient\quad =\quad 0,\\ f'(x)\quad =\quad 2x+6\quad =\quad 0\\ so,\quad 2x\quad =\quad -6\\ and\quad x\quad =\quad -3\)</span></p><p>c) <span class="math-tex">\(At\quad vertex,\quad f'(x)\quad =\quad 0\quad so\quad x\quad =\quad -3\\ Substitute\quad x\quad =\quad -3\quad into\quad f(x)={ x }^{ 2 }+6x+3\\ So,\quad y\quad =\quad -6\)</span> (You might use your table function here)</p></div><div class="actions"><span class="score" data-score="0"></span><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="label label-default q-number">3</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 3</p><p>Consider the function <span class="math-tex">\(f(x)=\frac { 1 }{ 3 } { x }^{ 3 }-\frac { 1 }{ 2 } { x }^{ 2 }-12x+4\)</span></p><p>Then <span class="math-tex">\(f'(x)=a{ x }^{ 2 }+b{ x }+c\)</span></p><p>a) What are the values of a, b and c?</p><p>b) What are the values of x when the gradient is zero? (Give your answers in numerical order)</p><p>c) What are the coordinates of the local maximum and minimum for this function</p></div><div class="q-answer"><p>a) a = <input type="text" style="height: auto;" data-c="1"> <span class="review"></span> , b = <input type="text" style="height: auto;" data-c="-1"> <span class="review"></span> , c = <input type="text" style="height: auto;" data-c="-12"> <span class="review"></span> </p><p>b) x = <input type="text" style="height: auto;" data-c="-3"> <span class="review"></span> , x = <input type="text" style="height: auto;" data-c="4"> <span class="review"></span> </p><p>c) Local maximum x = <input type="text" style="height: auto;" data-c="-3"> <span class="review"></span> , y = <input type="text" style="height: auto;" data-c="26.5"> <span class="review"></span> (give y to 3sf)</p><p>Local Minimum, x = <input type="text" style="height: auto;" data-c="4"> <span class="review"></span> , y = <input type="text" style="height: auto;" data-c="-30.7"> <span class="review"></span> (give y to 3sf)</p></div><div class="q-explanation"><p>a) <span class="math-tex">\(f'(x)={ x }^{ 2 }-{ x }-12\)</span>(Pay attention to the negative signs)</p><p>b) <span class="math-tex">\(When\quad the\quad gradient\quad is\quad zero\\ f'(x)={ x }^{ 2 }-{ x }-12\quad =\quad 0\\ f'(x)=(x+3)(x-4)\quad =\quad 0\\ x\quad =\quad -3\quad or\quad 4\)</span> You can use the polynomial solver on your GDC to solve this equation</p><p>c) Read the gradient function from your GDC for x = -3 and x = 4</p></div><div class="actions"><span class="score" data-score="0"></span><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="label label-default q-number">4</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 4</p><p>Consider the function <span class="math-tex">\(f(x)={ 2x }^{ 2 }+\frac { 32 }{ x } \)</span></p><p>a) This can be written as, <span class="math-tex">\(f(x)={ 2x }^{ 2 }+m{ x }^{ n }\)</span>, what are the values of m and n</p><p>b) The derivative of the function is given by <span class="math-tex">\(f'(x)\quad =\quad ax+b{ x }^{ c }\)</span>what are the values of a, b and c?</p><p>c) At what value of x is the gradient = 0?</p><p>d) What are the coordiantes of this local minimum?</p></div><div class="q-answer"><p>a) m = <input type="text" style="height: auto;" data-c="32"> <span class="review"></span> , n = <input type="text" style="height: auto;" data-c="-1"> <span class="review"></span> </p><p>b) a = <input type="text" style="height: auto;" data-c="4"> <span class="review"></span> , b = <input type="text" style="height: auto;" data-c="-32"> <span class="review"></span> , c = <input type="text" style="height: auto;" data-c="-2"> <span class="review"></span> </p><p>c) x = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> </p><p>d) x = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> , y = <input type="text" style="height: auto;" data-c="24"> <span class="review"></span> </p></div><div class="q-explanation"><p>a) BEWARE you are NOT differentiating here, you are just expressing it as a negative index. Remember...</p><p><span class="math-tex">\(\frac { 1 }{ { x }^{ n } } ={ x }^{ -n }\quad and\quad \frac { 3 }{ { x }^{ n } } ={ 3x }^{ -n }\quad and\quad \frac { 2 }{ { 5x }^{ n } } =\frac { 2 }{ 5 } { x }^{ -n }\)</span> so <span class="math-tex">\(f(x)={ 2x }^{ 2 }+\frac { 32 }{ x } \quad =\quad f(x)={ 2x }^{ 2 }+32{ x }^{ -1 }\)</span></p><p>...</p><p>b) <span class="math-tex">\(if,\quad f(x)={ 2x }^{ 2 }+32{ x }^{ -1 },\quad then,\quad f'(x)\quad =\quad 4x-32{ x }^{ -2 }\)</span></p><p>...</p><p>c) <span class="math-tex">\(when\quad gradient\quad =\quad 0,\\ f'(x)\quad =\quad 4x-32{ x }^{ -2 }\quad =\quad 0\\ f'(x)\quad =\quad 4x-\frac { 32 }{ { x }^{ 2 } } =0\\ so\quad 4x\quad =\quad \frac { 32 }{ { x }^{ 2 } } \\ and,\quad 4{ x }^{ 3 }\quad =\quad 32\\ and\quad { x }^{ 3 }\quad =\quad 8,\quad \\ so\quad x\quad =\quad 2\)</span> Again, your equation solver on the GDC can do this for you.</p><p>...</p><p>d) <span class="math-tex">\(At\quad vertex,\quad f'(x)\quad =\quad 0\quad so\quad x\quad =\quad 2\\ Substitute\quad x\quad =\quad 2\quad into\quad f(x)={ 2x }^{ 2 }+\frac { 32 }{ x } \\ So,\quad y\quad =\quad 24\)</span> Again, the table function will help you to avoid mistakes</p></div><div class="actions"><span class="score" data-score="0"></span><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="label label-default q-number">5</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 5</p><p>The Volume of water (<em>V</em>) in a container varies with some given conditions against time (<em>t</em>), The Volume, <span class="math-tex">\(V{ cm }^{ 3 }\)</span> , is given by <span class="math-tex">\(V=400+3t-{ t }^{ 2 }\)</span></p><p>a) If <span class="math-tex">\(\frac { dV }{ dt } =at+b\)</span> then what are the values of a and b?</p><p>b) At what value of t will the volume be at a maximum?</p><p>c) What will the maximum volume be?</p></div><div class="q-answer"><p>a) a = <input type="text" style="height: auto;" data-c="-2"> <span class="review"></span> , b = <input type="text" style="height: auto;" data-c="3"> <span class="review"></span> </p><p>b) t = <input type="text" style="height: auto;" data-c="1.5"> <span class="review"></span> (give the exact answer as a decimal)</p><p>c) Maximum Volume = <input type="text" style="height: auto;" data-c="402"> <span class="review"></span> </p></div><div class="q-explanation"><p>a) <span class="math-tex">\(if\quad V=400+3t-{ t }^{ 2 }\quad then,\frac { dV }{ dt } =3-2t\)</span></p><p>b) <span class="math-tex">\(at\quad maximum,\quad \frac { dV }{ dt } =3-2t=0\\ so,\quad 3=2t\\ and\quad t=\frac { 3 }{ 2 } =1.5\)</span></p><p>c) <span class="math-tex">\(Maximum\quad occurs\quad at\quad t\quad =\quad 1.5\\ when\quad t=1.5\\ V=400+3(1.5)-{ (1.5) }^{ 2 }\\ V=400+4.5-2.25\\ V=402.25,\quad V=402{ cm }^{ 3 }\quad (3sf)\)</span> </p><p>Using the GDC, if you enter the Volume function and have the derivative switched on then you can see that at t = 1.5, Volume is 402.25 and the derivative is zero.</p><p><img alt="" src="../../files/CLS/spo/q5.jpg" style="width: 302px; height: 141px;"></p></div><div class="actions"><span class="score" data-score="0"></span><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="label label-default q-number">6</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 6</p><p>A company's profit, P, in 1000s of euros can be modelled by the the function, <span class="math-tex">\(P(x)=19x-0.07{ x }^{ 2 }-10\)</span>where <span class="math-tex">\(x\)</span>, is a measure of units of product for the company.</p><p>a) if <span class="math-tex">\(\frac { dP}{ dx } =ax+b,\)</span>what are the values of a and b?</p><p>b) At what value of x while the profit be at a maximum?</p><p>c) Use the rounded answer to part b) to calculate the value of the maximum profit.</p></div><div class="q-answer"><p>a) a = <input type="text" style="height: auto;" data-c="-0.14"> <span class="review"></span> , b = <input type="text" style="height: auto;" data-c="19"> <span class="review"></span> </p><p>b) x = <input type="text" style="height: auto;" data-c="136"> <span class="review"></span> (give your answer to 3sf)</p><p>c) Maximum profit, P = <input type="text" style="height: auto;" data-c="1280"> <span class="review"></span> (give your answer as the number of 1000s to 3sf)</p></div><div class="q-explanation"><p>a) <span class="math-tex">\(if\quad P(x)=19x-0.07{ x }^{ 2 }-10,\quad then,\frac { dP }{ dx } =19-0.14x\)</span></p><p>b) <span class="math-tex">\(At\quad maximum,\quad \frac { dP }{ dx } =19-0.14x=0\\ 19=0.14x\\ x=\frac { 19 }{ 0.14 } =135.71428.....\\ x=\quad 136\quad (3sf)\)</span></p><p>c) <span class="math-tex">\(maximum\quad profit\quad occurs\quad at\quad x\quad =\quad 136\\ Substituting\quad in\quad to\quad P(x)=19x-0.07{ x }^{ 2 }-10\\ gives\quad V=1279.2\\ V=1280\quad (3sf)\)</span></p></div><div class="actions"><span class="score" data-score="0"></span><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="label label-default q-number">7</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 7</p><p>Consider the formula A = bh which is subject to the constraint that b - h=10. </p><p>a) Which of the expressions is the correct expression for A in terms of b?</p><p> <label class="radio"><input type="radio"> <span class="math-tex">\(A=bh\)</span></label> <label class="radio"><input class="c" type="radio"> <span class="math-tex">\(A={ b }^{ 2 }-10b\)</span></label> <label class="radio"><input type="radio"> <span class="math-tex">\(A=10{ b }^{ 2 }-b\)</span></label> <label class="radio"><input type="radio"> <span class="math-tex">\(A=10{ b }\)</span></label></p><p>b) If <span class="math-tex">\(\frac { dA }{ db } =mb+n\)</span>, what are the values of m and n?</p><p>c) For what value of b is A a minimum?</p><p>d) What is the minimum value of A?</p></div><div class="q-answer"><p>b) m = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> , n = <input type="text" style="height: auto;" data-c="-10"> <span class="review"></span> </p><p>c) b = <input type="text" style="height: auto;" data-c="5"> <span class="review"></span> </p><p>d) Minimum value of A = <input type="text" style="height: auto;" data-c="-25"> <span class="review"></span> </p></div><div class="q-explanation"><p>a) <span class="math-tex">\(A=bh\\ b-h=10\quad so\\ b=10+h\quad and\\ b-10=h\\ Substitute\quad h\quad =\quad b-10\\ A=b(b-10)\\ A={ b }^{ 2 }-10b\)</span></p><p>b) <span class="math-tex">\(if\quad A={ b }^{ 2 }-10b\\ then\quad \frac { dA }{ db } =2b-10\)</span></p><p>c) <span class="math-tex">\(at\quad minimum\quad value,\quad \frac { dA }{ db } =2b-10=0\\ so\quad 2b=10\\ and\quad b=5\)</span></p><p>d) <span class="math-tex">\(Substitute\quad b=5\quad into\quad A={ b }^{ 2 }-10b\\ so\quad A=-25\)</span></p></div><div class="actions"><span class="score" data-score="0"></span><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="label label-default q-number">8</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 8</p><p>Consider the formula <span class="math-tex">\(A={ x }^{ 3 }+2xh\)</span>, subject to the condition that <span class="math-tex">\(h=\frac { 3 }{ { x }^{ 3 } } \)</span></p><p>a) A can be expressed in terms of x only like this, <span class="math-tex">\(A={ x }^{ 3 }+p{ x }^{ q }\)</span>, what are the values of p and q?</p><p>b) <span class="math-tex">\(\frac { dA }{ dx } =m{ x }^{ 2 }-n{ x }^{ -3 }\)</span>, what are the values of m and n?</p><p>c) For what value of x is A at a local minimum?</p></div><div class="q-answer"><p>a) p = <input type="text" style="height: auto;" data-c="6"> <span class="review"></span> , q = <input type="text" style="height: auto;" data-c="-2"> <span class="review"></span> </p><p>b) m = <input type="text" style="height: auto;" data-c="3"> <span class="review"></span> , n = <input type="text" style="height: auto;" data-c="12"> <span class="review"></span> </p><p>c) x = <input type="text" style="height: auto;" data-c="1.32"> <span class="review"></span> (Answer to 3sf)</p></div><div class="q-explanation"><p>a) Substitute the value for h in to the expression for A,</p><p><span class="math-tex">\(if\quad A={ x }^{ 3 }+2xh\quad and\quad h=\frac { 3 }{ { x }^{ 3 } } \\ Substituting\quad gives\\ A={ x }^{ 3 }+\frac { 2x\times 3 }{ { x }^{ 3 } } \\ A={ x }^{ 3 }+\frac { 6x }{ { x }^{ 3 } } \\ A={ x }^{ 3 }+\frac { 6 }{ { x }^{ 2 } } \\ A={ x }^{ 3 }+6{ x }^{ -2 }\)</span></p><p>b) Differentiate A with respect to x</p><p><span class="math-tex">\(if\quad A={ x }^{ 3 }+6{ x }^{ -2 }\\ \\ then,\frac { dA }{ dx } =3{ x }^{ 2 }-12{ x }^{ -3 }\)</span></p><p>c) Solve for gradient = 0</p><p><span class="math-tex">\(At\quad minimum,\quad \frac { dA }{ dx } =3{ x }^{ 2 }-12{ x }^{ -3 }=0\\ 3{ x }^{ 2 }-12{ x }^{ -3 }=0\quad (use\quad GDC\quad solver)\\ x=1.3195....\\ x=1.32\quad (3sf)\)</span></p></div><div class="actions"><span class="score" data-score="0"></span><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="label label-default q-number">9</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 9</p><p>A closed cuboid box has Volume, <span class="math-tex">\(V={ x }^{ 2 }h\)</span>. Its surface area, A, is limted to <span class="math-tex">\(600{ cm }^{ 2 }\)</span></p><p><img alt="" src="../../files/CLS/spo/q8.jpg" style="width: 150px; height: 199px;">Dimensions shown are in cm</p><p>a) A is given by <span class="math-tex">\(A=m{ x }^{ 2 }+nxh\)</span>, what are the values of m and n?</p><p>b) Since A = 600, we can say that <span class="math-tex">\(\frac { a-{ x }^{ 2 } }{ bx } =h\)</span>, what are the values of a and b?</p><p>c) Use this expression for h to express the volume in terms of h only, where <span class="math-tex">\(V=px-\frac { 1 }{ 2 } { x }^{ q }\)</span>, what are the values of p and q?</p><p>d) For what value of x is the Volume at a maximum?</p><p>e) What is the maximum volume?</p><p>f) What will the value of h be at this volume?</p></div><div class="q-answer"><p>a) m = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> , n = <input type="text" style="height: auto;" data-c="4"> <span class="review"></span> </p><p>b) a = <input type="text" style="height: auto;" data-c="300"> <span class="review"></span> , b = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> </p><p>c) p = <input type="text" style="height: auto;" data-c="150"> <span class="review"></span> , q = <input type="text" style="height: auto;" data-c="3"> <span class="review"></span> </p><p>d) x = <input type="text" style="height: auto;" data-c="10"> <span class="review"></span> cm</p><p>e) Maximum Volume = <input type="text" style="height: auto;" data-c="1000"> <span class="review"></span> <span class="math-tex">\({ cm }^{ 3 }\)</span></p><p>f) h = <input type="text" style="height: auto;" data-c="10"> <span class="review"></span> cm</p></div><div class="q-explanation"><p>a) <span class="math-tex">\(Area\quad is\quad the\quad 6\quad faces\quad added\quad together\\ A\quad =\quad 2{ x }^{ 2 }+2xh+2xh\\ A=2{ x }^{ 2 }+4xh\)</span></p><p>b) This part is tricky but is all about making h the subject of the formula</p><p> <span class="math-tex">\(A=2{ x }^{ 2 }+4xh\\ 600=2{ x }^{ 2 }+4xh\\ 600-2{ x }^{ 2 }=4xh\\ \frac { 600-2{ x }^{ 2 } }{ 4x } =h\\ \frac { 300-{ x }^{ 2 } }{ 2x } =h\)</span></p><p>c) This stage involves substitutuing the expression for h in to the expression for volume.</p><p><span class="math-tex">\(Since\quad V{ =x }^{ 2 }h\quad and\quad h=\frac { 300-{ x }^{ 2 } }{ 2x } \\ V=\frac { { x }^{ 2 }(300-{ x }^{ 2 }) }{ 2x } \\ V=\frac { 300{ x }^{ 2 }-{ x }^{ 4 } }{ 2x } \\ (Divide\quad through\quad by\quad 2x)\\ V=150x-\frac { 1 }{ 2 } { x }^{ 3 }\)</span></p><p>d) Differentiate the function and solve for gradient = 0</p><p><span class="math-tex">\(if\quad V=150x-\frac { 1 }{ 2 } { x }^{ 3 }\quad then,\frac { dV }{ dx } =150-\frac { 3 }{ 2 } { x }^{ 2 }\\ Volume\quad is\quad a\quad maximum\quad when\quad \frac { dV }{ dx } =150-\frac { 3 }{ 2 } { x }^{ 2 }=0\\ 150=\frac { 3 }{ 2 } { x }^{ 2 }\\ 300=3{ x }^{ 2 }\\ 100={ x }^{ 2 }\\ x=\quad 10\quad (-10\quad is\quad not\quad in\quad the\quad valid\quad range)\)</span></p><p>e) This involves substituting the value of x that gives a minimum into the expression for volume.</p><p><span class="math-tex">\(Maximum\quad occurs\quad at\quad x=\quad 10\\ substitute\quad x=10\quad into\quad V=150x-\frac { 1 }{ 2 } { x }^{ 3 }\\ V=1000{ cm }^{ 3 }\)</span></p><p>f) Now substitute the value of x in to the expression for h.</p><p><span class="math-tex">\(Since\quad \frac { 300-{ x }^{ 2 } }{ 2x } =h\quad and\quad x=10\\ then,\quad \frac { 300-100 }{ 20 } =h\\ 10=h\)</span></p></div><div class="actions"><span class="score" data-score="0"></span><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="label label-default q-number">10</div><div class="exercise shadow-bottom"><div class="q-question"><p>Question 10</p><p>Consider a closed cylinder, radius r, height, h, whose volume must be 400<span class="math-tex">\({ cm }^{ 3 }\)</span>. The aim is to minimise the Surface area, A, required to make the cylinder.</p><p>a) if the height h, can be expressed as <span class="math-tex">\(h=\frac { a }{ \pi { r }^{ b } } \)</span>, what are the values of a and b?</p><p>b) The surface area of the cylinder can be expressed as <span class="math-tex">\(A=2\pi { r }^{ 2 }+p{ r }^{ q }\)</span>, what are the values of p and q?</p><p>c) What value of r will give the minimum surface area?</p><p>d) Use the rounded answer to the part c) to work out,</p><p>i) The value of h when A is a minimum</p><p>ii) The minimum value of A</p></div><div class="q-answer"><p>a) a = <input type="text" style="height: auto;" data-c="400"> <span class="review"></span> , b = <input type="text" style="height: auto;" data-c="2"> <span class="review"></span> </p><p>b) p = <input type="text" style="height: auto;" data-c="800"> <span class="review"></span> , q = <input type="text" style="height: auto;" data-c="-1"> <span class="review"></span> </p><p>c) r = <input type="text" style="height: auto;" data-c="ANSWER"> <span class="review"></span> (Answer correct to 3sf)</p><p>d) i) h = <input type="text" style="height: auto;" data-c="8.00"> <span class="review"></span> cm , ii) A = <input type="text" style="height: auto;" data-c="301"> <span class="review"></span> cm<sup>2</sup> (answers to 3sf)</p></div><div class="q-explanation"><p>a) Use the formula for volume of a cylinder and rearrange.</p><p><span class="math-tex">\(V=\pi { r }^{ 2 }h\quad and\quad is\quad given\quad as\quad 400{ cm }^{ 3 }\\ then\\ 400=\pi { r }^{ 2 }h\\ and\quad so,\quad h=\frac { 400 }{ \pi { r }^{ 2 } } \)</span></p><p>b) Use the formula for surface area of a cylinder and substitute the expression for h, then simplify.</p><p><span class="math-tex">\(A=2\pi { r }^{ 2 }+2\pi { r }h\quad and\quad h=\frac { 400 }{ \pi { r }^{ 2 } } \\ Substituting\quad for\quad h,\\ A=2\pi { r }^{ 2 }+\frac { 2\pi { r }\times 400 }{ \pi { r }^{ 2 } } \\ Cancelling\quad common\quad factors\\ A=2\pi { r }^{ 2 }+\frac { 800 }{ { r } } \\ A=2\pi { r }^{ 2 }+800{ r }^{ -1 }\)</span></p><p>c) Solve for gradient = 0. Perfectly acceptable to use the solve function to do this.</p><p><span class="math-tex">\(If\quad A=2\pi { r }^{ 2 }+800{ r }^{ -1 },\quad then\quad \frac { dA }{ dr } =4\pi { r }-800{ r }^{ -2 }\\ At\quad minimum,\quad \frac { dA }{ dr } =4\pi { r }-800{ r }^{ -2 }=0\\ 4\pi { r }=800{ r }^{ -2 }\\ 4\pi { r }=\frac { 800 }{ { r }^{ 2 } } \\ 4\pi { r }^{ 3 }=800\\ { r }^{ 3 }=\frac { 800 }{ 4\pi } \\ r=\sqrt [ 3 ]{ \frac { 800 }{ 4\pi } } \\ r=3.9929454...\\ r=3.99\quad cm\quad (3sf)\)</span></p><p>d) Substitutions can be done using the table function on your GDC or otherwise.</p><p>i) <span class="math-tex">\(Substitute\quad r=3.99\quad in\quad to\quad h=\frac { 400 }{ \pi { r }^{ 2 } } \\ h=7.997685...\\ h=8.00cm\quad (3sf)\)</span></p><p>ii) <span class="math-tex">\(Substitute\quad r=3.99\quad in\quad to\quad A=2\pi { r }^{ 2 }+800{ r }^{ -1 }\\ A=300.5301915...\\ A=\quad 301{ cm }^{ 2 }\quad (3sf)\)</span></p></div><div class="actions"><span class="score" data-score="0"></span><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="totals"><span class="score"></span><button class="btn btn-success btn-block text-center check-total"><i class="fa fa-check-square-o"></i> Check</button></div></div><hr> </section> </div> </div> <div class="panel-footer"> <div> </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"> <p>The following questions are based on IB exam style questions from past exams. You should print these off (from the document at the top) and try to do these questions under exam conditions. Then you can check your work with the video solution.</p> <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>Question 1</p> </div> </div> <div class="panel-body"> <p><img alt="" src="../../files/CLS/spo/q1.jpg" style="width: 700px; height: 336px;"></p> <p>Video solution</p> <button class="btn btn-xs bg-turquoise showhider"><i class="fa fa-fw fa-plus"></i></button><section class="hiddenbox hidden"> <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/668972314"></iframe></div> <p>Worked Solution</p> <p><img alt="" src="../../files/CLS/spo/s1.jpg" style="width: 700px; height: 710px;"></p> </section> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-default 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