Science:Math Exam Resources/Courses/MATH100/December 2011/Question 06

MATH100 December 2011

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Question 06

Full-Solution Problems. In questions 2-8, justify your answers and show all your work. Simplification of answers is not required unless explicitly stated. If 24m2 of material is available to make a rectangular storage container with an open top, and if the length of its base is twice the width, find the largest possible volume of the rectangular storage container. Please justify that your answer gives indeed the largest possible 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 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
Try drawing a picture. The object we are looking for is a rectangular prism with an open top. What is the volume of such an object? What is the surface area?

Hint 2
Setting l to be the length, w to be the width and h to be the height, we have ${\begin{aligned}V&=lwh\\SA&=2lh+2wh+lw\\l&=2w\end{aligned}}$ where for the surface area above, we keep in mind that our box has no top. Since this is an optimization problem, which function are you trying to maximize? Which functions are the constraints?

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. We will need the following formulas: The volume of the box is given by $\displaystyle V=lwh$ Where l is the length of the base of the box, w is the width, and h is the height. The material will form the outside of the box, which is the same as its surface area. The surface area of an open box is given by $\displaystyle SA=2lh+2wh+lw$ With l, w, and h as before. We also know that the length of the base of the box is twice its width. In an equation, this means $\displaystyle l=2w$ We can plug this into our volume and surface area formulas to get $\displaystyle V=2w^{2}h$ $\displaystyle SA=6wh+2w^{2}$ We know one more fact. Because we have $24m^{2}$ of material, the formula for surface area must be equal to 24, or $\displaystyle 24=6wh+2w^{2}$ Isolating for h yields $\displaystyle h={\frac {24-2w^{2}}{6w}}$ (NOTE: In this step we are assuming that the width is not equal to zero, which makes sense given the physical limitations of our problem, that is, boxes have positive dimensions). Plugging this into our formula for volume yields $\displaystyle V=2w^{2}h=2w^{2}\cdot {\frac {24-2w^{2}}{6w}}={\frac {w(24-2w^{2})}{3}}=8w-{\frac {2}{3}}w^{3}$ We wish to optimize V. Now that is has only one variable, differentiating yields $\displaystyle V'=8-2w^{2}$ Setting this equal to 0 and solving gives $w^{2}=4$ and thus that $w=2$ (where we note again that dimensions must be positive). Looking at the derivative of volume, a sign chart tells us that$V'(w)>0$ when $w<2$ and $V'(w)>0$ when $w<2$. Hence, we have a maximum at w=2. The volume of the box with this width is $\displaystyle V(2)=8(2)-{\frac {2}{3}}(2)^{3}=16-{\frac {16}{3}}={\frac {32}{3}}$ and this is the maximal volume as required.

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