Question

An open-top box with a square base has a surface area of $1200{\text{ }}square{\text{ }}inches$. Find the largest possible volume of the box?

Answer 1

Given problem tests the concepts of derivatives and their applications. These type questions can be easily solved if we keep in mind the concepts of maxima and minima. In the problem, we are required to find the largest possible volume of the box having surface area $1200{\text{ }}square{\text{ }}inches$ . We have to first find the volume of the open square box as a function of the known quantity and then differentiate it to find the points of local maxima or minima.

Complete step by step answer:
Surface area of open top box with square base $= 1200{\text{ }}square{\text{ }}inches - - - - - - - (1)$

Let each side of base $= x{\text{ inches}}$

And height of the box $= h{\text{ inches}}$

So, Surface area of open top box with square base = area of 4 walls + area of base

Surface area of open top box with square base = $4h(x + x) + {x^2}$

Surface area of open top box with square base = $4h\left( {2x} \right) + {x^2}$

Surface area of open top box with square base = $8hx + {x^2}$

So, $8hx + {x^2} = 1200$square inches, using equation (1)
=$h = \left( {\dfrac{{1200 - {x^2}}}{{8x}}} \right) - - - - - - (2)$

Now for largest possible volume:
Volume of open top box with square base = area of base x height

Volume of open top box with square base $= {x^2} \times h$

Volume of open top box with square base $= {x^2} \times \left[ {\dfrac{{1200 - {x^2}}}{{8x}}} \right]$, using equation (2)

Volume of open top box with square base $= x \times \left[ {\dfrac{{1200 - {x^2}}}{8}} \right]$

Volume of open top box with square base $= \dfrac{1}{8}\left[ {1200x - {x^3}} \right]$

Now consider volume as function of ‘x’ write above expression
$V(x) = \dfrac{1}{8}\left[ {1200x - {x^3}} \right]$

Differentiating with respect to x on both sides
$V'(x) = \dfrac{1}{8}\left[ {1200 - 3{x^2}} \right] - - - - - - (3)$

Differentiating with respect to x on both sides to get second order derivative,
$V''(x) = \dfrac{1}{8}\left[ {0 - 6x} \right]$
Or $V''(x) = - \dfrac{{3x}}{4}\; - - - - (4)$

Now to get critical points or values of x, consider $V'(x) = 0$
From equation (3)

$\dfrac{1}{8}\left[ {1200 - 3{x^2}} \right] = 0$
Or $3{x^2} - 1200 = 0$
Or ${x^2} = 400$
Or $x = \pm 20$

Taking only $x = 20$ and substituting in equation (4), $x = \left( { - 20} \right)$ cannot be taken into consideration as measure of a side can never be negative.
$V''(x) = - \dfrac{{60}}{4}\; = - 15$

Since$V''(x)$is negative at $x = 20$, so by the second order derivative test, Volume of box is largest when $x = 20$ .

Now substitute $x = 20$in equation (2) to get value of ‘h’
$h = \dfrac{{1200 - 400}}{{160}} = \dfrac{{800}}{{160}} = 5$

Thus for largest volume $x = 20$and $h = 5$

Therefore largest volume of the given box $= {(20)^2} \times 5{\text{ }}cubic{\text{ }}inches$
Or largest volume of the given box$= 2000{\text{ }}cubic{\text{ }}inches$

Note: We can solve the problems involving the maxima and minima concept by two methods: the first derivative test and the second derivative test. The second derivative test involves finding the local extremum points and then finding out which of them is local minima and which is local maxima.