Science:Math Exam Resources/Courses/MATH104/December 2016/Question 12

MATH104 December 2016

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[hide] Question 12
Suppose you are given a 12" x 12" square piece of cardboard and asked to construct a box by cutting out squares of equal size from the four corners and bending up the sides. Find the dimensions of the resulting box that has the largest volume. Explain carefully how you may conclude that your answer is guaranteed to be a box of maximal volume.

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!

[show] Hint
Set $x$ as the side length of the square which we cut out from the four corner. Express the volume of the resulting box (described in the question) as a function of $x$. Determine the domain of the function and find its global maximum on the domain.

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[show] 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. Let's begin with diagram: Let $x\geq 0$ be the side length of the square which we cut out from the four corner. Indeed, $x$ becomes the height of the resulting box, The bottom of the box is also a square with the side length $12-2x\geq 0$ as we can see in the diagram. Then, the volume of the box is $$V=x(12-2x)^{2}$$ Since $x>0$ and $12-2x>0$ is required, the domain of the function $V$ is $0\leq x\leq 6$. To find the largest volume (which is the global maximum of $V$ on $[0,6]$, we first find the derivative of $V$; using the product rule and the chain rule, $${\begin{aligned}{\frac {dV}{dx}}&=(12-2x)^{2}+x\cdot 2(12-2x)(-2)\\&=(12-2x)[(12-2x)-4x]\\&=(12-2x)(12-6x)=12(6-x)(2-x)\end{aligned}}$$ Setting ${\textstyle {\frac {dV}{dx}}=0}$ gives us the critical numbers: ${\textstyle x=6}$ and ${\textstyle x=2}$. Since we have a closed interval, we can test the critical numbers and the 2 endpoints. $${\begin{aligned}V(0)&=0\\V(2)&=2\cdot 8^{2}\\V(6)&=0\end{aligned}}$$ From which we see there is global maximum when ${\textstyle x=2}$. The height of the resulting box will be ${\textstyle x=2}$, while the length and width will be ${\textstyle 12-2x=8}$, therefore the dimensions are $\color {blue}8\times 8\times 2$.