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

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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 $y=9-x^{2}.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

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
By drawing the graph of the parabola and the rectangle, the graph of y=9-x^2 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 $\color {blue}2{\sqrt {3}},6$ .