Question

What is the maximum volume of an open box with a square base whose surface area (not including the top) is 27 $\text{i}{\text{n}}^2$?

Answer 1

Hint: Let us assume an open box, with a square base with each edge of length a inch, and height of the box h inch.

Let the surface area of the box (not including the top the top) be s, then we have
$s=(a\times a)+4(a\times h)$
Thus, the surface area of the open box is $s=a^2+4ah$.
We can also write the above equation as
 $4ah=s-a^2$
And thus, we have the height of the open box as
$h={\displaystyle \frac{s-a^2}{4a}}...\left(i\right)$

Complete step-by-step solution:
From the figure, we can clearly say that the volume of this open box is
$V=a\times a\times h$
Using the value of height from equation (i), we get
$V=a^2\left({\displaystyle \frac{s-a^2}{4a}}\right)$
We can simplify this equation by cancelling terms to get,
$V=\left({\displaystyle \frac{as-a^3}{4}}\right)...\left(ii\right)$
Here, we know that the value of s is constant, and we need to find the value of a for which the volume is maximum.
So, we know that we need to differentiate the volume with respect to a. Thus, we have
${\displaystyle \frac{dV}{da}}={\displaystyle \frac{d}{da}}\left({\displaystyle \frac{as-a^3}{4}}\right)$
So, on differentiation, we get
${\displaystyle \frac{dV}{da}}={\displaystyle \frac{d}{da}}\left({\displaystyle \frac{as}{4}}\right)-{\displaystyle \frac{d}{da}}\left({\displaystyle \frac{a^3}{4}}\right)$
Hence, we now have the following equation
${\displaystyle \frac{dV}{da}}={\displaystyle \frac{s}{4}}-{\displaystyle \frac{3a^2}{4}}$
We now need to equate this differentiation with zero. So, we get
${\displaystyle \frac{s}{4}}-{\displaystyle \frac{3a^2}{4}}=0$
Solving for a, we get
$3a^2=s$
Or, we can write
$a=\sqrt{{\displaystyle \frac{s}{3}}}$
We still need to see whether this is a maxima point or minima point.
So, ${{\displaystyle \frac{d^{2}V}{d{a}^2}}|}_{a=\sqrt{{\displaystyle \frac{s}{3}}}}={\displaystyle \frac{d}{da}}{[{\displaystyle \frac{s}{4}}-{\displaystyle \frac{3a^2}{4}}]}_{a=\sqrt{{\displaystyle \frac{s}{3}}}}$
On solving this equation, we get
${{\displaystyle \frac{d^{2}V}{d{a}^2}}|}_{a=\sqrt{{\displaystyle \frac{s}{3}}}}={[0-{\displaystyle \frac{3a}{2}}]}_{a=\sqrt{{\displaystyle \frac{s}{3}}}}$
On putting the value of a, we get
${{\displaystyle \frac{d^{2}V}{d{a}^2}}|}_{a=\sqrt{{\displaystyle \frac{s}{3}}}}=[-{\displaystyle \frac{3\sqrt{{\displaystyle \frac{s}{3}}}}{2}}]$
Or, we can write
${{\displaystyle \frac{d^{2}V}{d{a}^2}}|}_{a=\sqrt{{\displaystyle \frac{s}{3}}}}=[-{\displaystyle \frac{\sqrt{3s}}{2}}]$
We can clearly see that ${{\displaystyle \frac{d^{2}V}{d{a}^2}}|}_{a=\sqrt{{\displaystyle \frac{s}{3}}}}<0$ .
Thus, the volume is maximum at $a=\sqrt{{\displaystyle \frac{s}{3}}}$ .
We know that the surface area is 27 $\text{i}{\text{n}}^2$ as is given in the question. So, we now get the value
$a=\sqrt{{\displaystyle \frac{27}{3}}}$
And since the value of a can never be negative, the value of a will be
$a=3$ .
Putting the values of a and s in equation (ii), we get
$V=\left({\displaystyle \frac{(3\times 27)-{\left(3\right)}^3}{4}}\right)$
Thus, we get
$V=\left({\displaystyle \frac{(3\times 27)-27}{4}}\right)$
On solving the above equation, we get
$V=\left({\displaystyle \frac{27}{2}}\right)\text{i}{\text{n}}^3$
Hence, $V=13.5\text{ i}{\text{n}}^3$
Thus, the maximum volume of the open box is $13.5\text{ i}{\text{n}}^3$.

Note: We must remember that the surface area of open box with a square base is $s=a^2+4ah$ and the surface area of a closed box with a square base is $s=2a^2+4ah$. Here, $a=\sqrt{{\displaystyle \frac{s}{3}}}$ , and since a can not be a negative value, we can say that $\sqrt{s}$ is also positive and thus, $[-{\displaystyle \frac{\sqrt{3s}}{2}}]$ will be negative.