Science:Math Exam Resources/Courses/MATH104/December 2015/Question 09/Solution 1

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 ${\color {blue}2{\text{ m}}\times 2{\text{ m}}\times 2{\text{ m}}}.$