Question

If $40$ square feet of sheet metal are to be used in the construction of an open tank with square base, find the maximum capacity of the tank.

• A. $\frac{40}{3}\sqrt{\frac{10}{3}}$
• B. $\frac{20}{3}\sqrt{\frac{10}{3}}$
• C. $\frac{10}{3}\sqrt{\frac{10}{3}}$
• D. $\frac{10}{3}\sqrt{\frac{20}{3}}$

Answer 1

The correct option is A $\frac{40}{3}\sqrt{\frac{10}{3}}$

Let the dimension of the tank with square base be $x,x,h$
$x^2+4hx=40$
$\Rightarrow h=\frac{40-x^2}{4x}$
$V=x^{2}h$
$\Rightarrow V=x\left(\frac{40-x^2}{4}\right)$
$\Rightarrow V=\frac{40x-x^3}{4}$ ....(2)
For maxima or minima,
$\frac{dV}{dx}=0$
$\Rightarrow 40-3x^2=0$
$x=2\sqrt{\frac{10}{3}}$
Now, $\frac{d^{2}V}{d{x}^2}<0$
Hence, $V$ is maximum at $x=2\sqrt{\frac{10}{3}}$
By (1), $h=\sqrt{\frac{10}{3}}$
So, maximum volume $=\frac{40}{3}\sqrt{\frac{10}{3}}$