Science:Math Exam Resources/Courses/MATH104/December 2015/Question 09

MATH104 December 2015

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Question 09
A carpenter has been asked to build an open box with a square base, where an open box means a box without a top. The sides of the box will cost $2.50 per square metre and the base will cost $5 per square metre. What are the dimensions of the box of maximal volume that can be constructed for $60? Note: In order to receive full credit, you will need to 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
Let $\ell$ be the side length of the box's (square) base, and let $h$ be its height. Can you find an expression for the volume of the box in terms of $\ell$ and $h$ (which we will subsequently seek to maximize)?

Hint 2
Continuing from the first hint, we know that the total cost of the box must be $60. Given that the sides of the box cost $2.50/m2 and the base costs $5/m2, can you express this constraint as an equation involving $\ell$ and $h$? (First try finding expressions for the areas of the sides and base.)

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. Let $\ell$ and $h$ be the base side length and height of the box, respectively (both in metres). The volume of the box is $\ell \cdot \ell \cdot h=\ell ^{2}h$ cubic metres. The cost of the box base is $\${\color {OliveGreen}5}$ per square metre; since the area of the base is $\ell \cdot \ell =\ell ^{2}$ square metres, the cost of the box base is ${\color {OliveGreen}5}\ell ^{2}$ dollars. The cost of the box sides is $\${\color {OrangeRed}2.50}$ per square metre; since there are four sides to the open top box, each with area $\ell \cdot h$ square metres, the cost of the box sides is ${\color {OrangeRed}2.50}\cdot 4\ell h=10\ell h$ dollars. We are given that the total cost of the box is $\$60,$ so our full optimization problem is ${\begin{aligned}{\text{maximize}}&&&\ell ^{2}h\\{\text{subject to}}&&&5\ell ^{2}+10\ell h=60\,.\end{aligned}}$ Solving for $h$ in the constraint equation, we obtain ${\begin{aligned}5\ell ^{2}+10\ell h&=60\\10\ell h&=60-5\ell ^{2}\\h&={\frac {60-5\ell ^{2}}{10\ell }}={\color {Orange}{\frac {6}{\ell }}-{\frac {\ell }{2}}}\,.\end{aligned}}$ Now substituting this into into the objective function, we have ${\begin{aligned}{\text{maximize}}&&&\ell ^{2}\left({\color {Orange}{\frac {6}{\ell }}-{\frac {\ell }{2}}}\right)=6\ell -{\frac {\ell ^{3}}{2}}\,.\end{aligned}}$ If we let $V(\ell )=6\ell -{\frac {\ell ^{3}}{2}},$ we can maximize $V$ over $\ell$ by setting $V'(\ell )=0$ and finding critical points: ${\begin{aligned}V'(\ell )&=6-{\frac {3}{2}}\ell ^{2}\\0&=6-{\frac {3}{2}}\ell ^{2}\\{\frac {3}{2}}\ell ^{2}&=6\\\ell &={\sqrt {4}}=\pm 2\,.\end{aligned}}$ But since we must have $\ell >0$ (i.e., the side length of the box must be positive), it is $\ell =2$ which maximizes the volume of the box. We verify that this is indeed a maximum by taking the second derivative of $V$ and evaluating it at $\ell =2:$ $V''(2)=\left.-3\ell \right|_{\ell =2}=-6<0.$ Hence by the second derivative test, $V$ attains a local maximum when $\ell =2.$ Since the height of the box is $h={\tfrac {6}{\ell }}-{\tfrac {\ell }{2}},$ when $\ell =2$ we have $h={\tfrac {6}{2}}-{\tfrac {2}{2}}=2.$ Therefore the dimensions of the box of maximal volume are ${\color {blue}2{\text{ m}}\times 2{\text{ m}}\times 2{\text{ m}}}.$