A metal box with a square base and vertical sides is to contain 1024…

The correct answer is Option (2) → ₹1920

Let x cm be the side of a square box and h cm be its height, then volume of box = $x^2h = 1024$ (given)

$⇒h =\frac{1024}{x^2}$   ...(i)

Let C (in ₹) be the cost of the box, then

C = 5 (area of top + area of bottom) + $\frac{5}{2}$ × area of sides

$=5 (x^2 + x^2) +\frac{5}{2}×4xh$

$=10x^2+10x×\frac{1024}{x^2}$   (using (i))

$=10\left(x^2+\frac{1024}{x}\right)$.

Differentiating (i) w.r.t. x, we get

$\frac{dC}{dx}=10\left(2x+\frac{1024}{x^2}\right)$ and $\frac{d^2C}{dx^2}=10\left(2+2.\frac{1024}{x^3}\right)$.

Now $\frac{dC}{dx}=0⇒10\left(2x+\frac{1024}{x^2}\right)=0⇒x-\frac{512}{x^2}=0$

$⇒x^3=512⇒x=8$

$\left(\frac{d^2C}{dx^2}\right)_{x=8}=10\left(2+2×\frac{1024}{512}\right)=60>0$

⇒ C is minimum when $x=8$.

Minimum cost = $₹10\left(64+\frac{1024}{8}\right)=₹10(64+128)=₹1920$