Science:Math Exam Resources/Courses/MATH110/April 2018/Question 09

MATH110 April 2018

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Question 09
Find the dimensions of the rectangle with the largest area that can be drawn with its base on the $x$ -axis and its upper corners on the parabola $y=9-x^{2}.$

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Hint 1
Find the length of the base and height of the rectangle which has its base on the $x$ -axis and its upper corners on the parabola $y=9-x^{2}.$

Hint 2
Find the area function of the rectangle and the range of $x$ for which the lengths of the base and height are positive.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. By drawing the graph of the parabola and the rectangle, we can see that the length of the the base is $2x$ and that of the height is $y=9-x^{2}$ . The base is positive if $x>0$ and the height is positive when $-3<x<3$ . (See the graph of the parabola.) Therefore, we consider $x$ on $(0,3)$ . Since the area of a rectangle is the length of its base times that of its height, the area function $A$ can be defined as $A(x)=2x\cdot (9-x^{2}).$ To find the maximum of $A$ on the interval $(0,3)$ , we find the critical points first. By the product rule for a derivative, the derivative of $A$ is $A'(x)=2(9-x^{2})+2x\cdot (-2x)=18-2x^{2}-4x^{2}=18-6x^{2}=6(3-x^{2})=6({\sqrt {3}}-x)({\sqrt {3}}+x).$ Then, solving $A'(x)=0$ , we get $x=-{\sqrt {3}},{\sqrt {3}}.$ Since $-{\sqrt {3}}$ is outside of the interval $(0,3)$ , we only consider ${\sqrt {3}}$ . Observing that $A'(x)=6({\sqrt {3}}-x)({\sqrt {3}}+x)>0$ on $(0,{\sqrt {3}})$ , $A$ is increasing on $(0,{\sqrt {3}})$ . On the other hand, since $A'(x)=6({\sqrt {3}}-x)({\sqrt {3}}+x)<0$ on $({\sqrt {3}},3)$ , $A$ is decreasing on $({\sqrt {3}},3)$ . Therefore, we obtain the maximum value of the area $A$ at $x={\sqrt {3}}$ . Calculating the length of the base and height at $x={\sqrt {3}}$ , we have ${\text{(the length of the base)}}=2{\sqrt {3}}$ and ${\text{(the length of the height)}}=9-({\sqrt {3}})^{2}=6$ . Answer: The length of the base and height are $\color {blue}2{\sqrt {3}},6$ .