Science:Math Exam Resources/Courses/MATH110/April 2016/Question 07

MATH110 April 2016

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[hide] Question 07
If $1200{\text{cm}}^{2}$ of material is available to make a box with a square base and open top, find the largest possible volume of the box. Justify your answer. Recall the volume $V$ of a rectangular box with width $w$, length $l$, and height h is $V=wlh$.

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

You want to maximize the volume of the box. Therefore, this is an optimization problem. Write down an equation for the volume of the box in terms of the side length $\ell$ of the base, and the height $h$ of the box. Then, write down an expression for the surface area of this box. There is a square base of side length $\ell$ and four rectangles making up the sides of the box. You know that the surface area should be equal to $1200$ square centimeters. You can use this to find a relationship between $\ell$ and $h$ to plug into your volume equation.

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[show] Solution

Note: We will carry the units through in all of our calculations. If you find this confusing, please ignore all of the units. Consider a square box with an open top. We will call the sidelength of the square base $\ell$ and call the height of the box $h$. We want to maximize the volume, $V$ of the box. A formula for the volume of this box is $V=\ell ^{2}h.$ We know that the surface area of the box should be $1200{\text{cm}}^{2}$. The box consists of a base of area $\ell ^{2}$ and four rectangles of length $\ell$ and height $h$. Therefore, the surface area of the box $\ell ^{2}+4\ell h$. Therefore, we have that $\ell ^{2}+4\ell h=1200{\text{cm}}^{2}.$ We will solve this equation for $h$. It would have also been possible to solve this equation for $\ell$, but solving for $h$ is simpler. We subtract $\ell ^{2}$ from both sides to get $4\ell h=1200{\text{cm}}^{2}-\ell ^{2}$ and then divide both sides by $4\ell$: $h={\frac {1200{\text{cm}}^{2}-\ell ^{2}}{4\ell }},$ and rewrite the right side of this equation as two separate fractions: $h={\frac {1200{\text{cm}}^{2}}{4\ell }}-{\frac {\ell ^{2}}{4\ell }}.$ Finally, we simplify the fractions: $h={\frac {300{\text{cm}}^{2}}{\ell }}-{\frac {\ell }{4}}.$ Now, we will plug this expression for $h$ into our volume equation : ${\begin{aligned}V&=\ell ^{2}h\\V&=\ell ^{2}\left({\frac {300{\text{cm}}^{2}}{\ell }}-{\frac {\ell }{4}}\right)\\V&={\frac {300\ell ^{2}{\text{cm}}^{2}}{\ell }}-{\frac {\ell ^{3}}{4}}\\V&=300\ell {\text{cm}}^{2}-{\frac {1}{4}}\ell ^{3}.\end{aligned}}$ We want to maximize $V$. Before proceeding with the optimization, we should identify the domain of our function. Clearly, $\ell$ must be at least 0. The largest value of $\ell$ occurs when the height is zero, in which case we have $\ell ^{2}=1200{\text{cm}}^{2}$, which means that $\ell ={\sqrt {1200}}{\text{cm}}$. Now that we have identified the domain, we should find the critical points of $V$. To do this, we need to compute the derivative of $V$. ${\frac {dV}{d\ell }}=300{\text{cm}}^{2}-{\frac {3}{4}}\ell ^{2}.$ In order to find the critical points, we set the derivative equal to zero: ${\begin{aligned}0&=300{\text{cm}}^{2}-{\frac {3}{4}}\ell ^{2}\\{\frac {3}{4}}\ell ^{2}&=300{\text{cm}}^{2}\\\ell ^{2}&=400{\text{cm}}^{2}\\\ell &=20{\text{cm}}\end{aligned}}.$ so the critical point occurs when $\ell =20$. We now plug in our critical points, as well as the endpoints of our domain, into the equation for $V$: $V(\ell )=300\ell {\text{cm}}^{2}-{\frac {1}{4}}\ell ^{3}$ When $\ell =0$, we get $V(0)=300(0){\text{cm}}^{2}-{\frac {1}{4}}0^{3}$ which is zero. When $\ell ={\sqrt {1200}}{\text{cm}}$, we get: $V({\sqrt {1200}}{\text{cm}})=300{\sqrt {1200}}{\text{cm}}^{3}-{\frac {1}{4}}*({\sqrt {1200}}{\text{cm}})^{3},$ but $({\sqrt {1200}})^{3}$ is $1200{\sqrt {1200}}$, so this is ${\begin{aligned}V({\sqrt {1200}}{\text{cm}})&=300{\sqrt {1200}}{\text{cm}}^{3}-{\frac {1200}{4}}{\sqrt {1200}}{\text{cm}}^{3}\\&=300{\sqrt {1200}}{\text{cm}}^{3}-300{\sqrt {1200}}{\text{cm}}^{3}\\&=0\end{aligned}}$ So, the only option left is $\ell =20{\text{cm}}$. When we plug this in, we get ${\begin{aligned}V(20{\text{cm}})&=300(20{\text{cm}}^{3})-{\frac {1}{4}}(8000{\text{cm}}^{3})\\&=6000{\text{cm}}^{3}-2000{\text{cm}}^{3}\\&=4000{\text{cm}}^{3}\\&=4\,{\text{liters}}\end{aligned}}$ So the maximum possible volume for the box is 4 liters, or 4000 cubic centimeters. Answer: $\color {blue}4000{\text{cm}}^{3}$