Science:Math Exam Resources/Courses/MATH100/December 2013/Question 09

MATH100 December 2013

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Question 09
Full-Solution Problem. Justify your answer and show your work. Full simplification of numerical answers is required unless explicitly stated otherwise. Find the area of the largest rectangle which has two vertices on the x-axis and two vertices lie on the graph of the function $\displaystyle y=8-x^{2}$ with $\displaystyle -{\sqrt {8}}\leq x\leq {\sqrt {8}}$. Please justify your answer.

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
Note that if two vertices of the rectangle must lie on the graph of the function $\displaystyle y=8-x^{2}$, the two vertices on the x-axis must be equidistant from the origin.

Hint 2
Suppose that the base of the rectangle measures $2b$ units long. What is an expression for the area of the rectangle, in terms of $b$?

Hint 3
If the base of the rectangle measures $2b$, its bottom right vertex is located at $(b,0)$. Therefore its top right vertex is located at $(b,8-b^{2})$ and it follows that its dimensions are $2b$ by $8-b^{2}$. Hence its area is $2b(8-b^{2})$. How can we maximize the value of this expression?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. If the base of the rectangle measures $\displaystyle 2b$, its area is $\displaystyle A(b)=2b(8-b^{2})=16b-2b^{3}$. To maximize $\displaystyle A$, we find its critical points by finding $\displaystyle A'$: ${\begin{aligned}A(b)&=16b-2b^{3}\\A'(b)&=16-6b^{2}\\\\0&=16-6b^{2}\\6b^{2}&=16\\b&=\pm {\sqrt {\frac {8}{3}}}\end{aligned}}$ A simple inspection shows that $b={\sqrt {\frac {8}{3}}}$ is a maximum point of A since A' changes sign from positive to negative at that point. The value of A(b) at that b (i.e., the maximum area) is: ${\begin{aligned}A\left({\sqrt {\frac {8}{3}}}\right)&=2{\sqrt {\frac {8}{3}}}\left(8-\left({\sqrt {\frac {8}{3}}}\right)^{2}\right)\\&={\color {blue}{\frac {32}{3}}\cdot {\sqrt {\frac {8}{3}}}}={\color {blue}{\frac {64{\sqrt {2}}}{3{\sqrt {3}}}}}={\color {blue}{\frac {64{\sqrt {6}}}{9}}}\end{aligned}}$

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