Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 12 (a)

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MATH 180 December 2017

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Other MATH 180 Exams

• December 2017 •

Question 12 (a)
Suppose you wish to construct an enclosed box of maximal volume out of an ${\textstyle L\times L}$ square sheet of cardboard by removing the shaded areas below and folding along the dotted lines. (Each unshaded region will form exactly one side of the box.) $A box$ A box (a) Write down an expression for the volume ${\textstyle V(x)}$ of the box.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Denote the length and width of the enclosed box by $l$ and $w$ , respectively. Then, using the given information, write them in terms of the height $x$ of the box and the length $L$ of the square sheet.

Checking a solution serves two purposes: helping you if, after having used the hint, 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.
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Solution
By the Hint, we first denote the length and width of the enclosed box by $l$ and $w$ , respectively. Its height is given by $x$ . Then, the volume of the box is $lwx$ . Now, we write $l$ and $w$ in terms of $x$ and $L$ . Since the width of the square sheet is same with the summation of double of height of the box and the length of the box, we get $L=2x+l$ . On the other hand, the height of the square sheet is same with the summation of double of height of the box and double of the width of the box, so that $L=2x+2w$ . Therefore, $l$ and $w$ can be expressed as $l=L-2x$ and $w={\frac {L}{2}}-x$ . Plugging these into the volume formula, we get $V(x)=lwx=(L-2x)\left({\frac {L}{2}}-x\right)x$ . Answer: $\color {blue}V(x)=lwx=(L-2x)\left({\frac {L}{2}}-x\right)x$