Question

A metal box with a square base and vertical sides is to contain $1024c{m}^3$. The material for the top and bottom costs $₹5/c{m}^2$ and the material for the sides costs $₹2.50/c{m}^2$. Find the least cost of the box.

Answer 1

Apply the concept of Maxima and Minima.

Given the volume of the box $=1024c{m}^3$

Let length of the side of square base be $xcm$ and height of the box be $ycm$.

Volume of box $=x^{2}y$

$$\begin{gathered} \Rightarrow x^{2}y=1024 \\ \Rightarrow y=\frac{1024}{x^2} \end{gathered}$$

Let $\text{'}C\text{'}$ be the cost of the material of the box.

So,

$$\begin{gathered} C=2x^2\times \left(5\right)+4xy\times 2.5 \\ =10x^2+10xy \\ =10x(x+y) \\ =10x\left(x+\frac{1024}{x^2}\right) \\ =10x^2+\frac{10240}{x} \end{gathered}$$

Differentiate both sides w.r.t. $x$

$\frac{dC}{dx}=20x-\frac{10240}{x^2}$

For maxima or minima, equating derivative to zero, we get

$$\begin{gathered} \frac{dC}{dx}=0\Rightarrow 20x-\frac{10240}{x^2}=0 \\ \Rightarrow 20x=\frac{10240}{x^2} \\ \Rightarrow x^3=\frac{1024}{2} \\ \Rightarrow x^3=512 \\ \Rightarrow x=8 \end{gathered}$$

Differentiating again for $x=8$, we get

$\frac{d^{2}C}{d{x}^2}=20+\frac{2\times 10240}{x^3}>0$

So, $x=8$ is a point of minima.

We have:

$C=10x^2+\frac{10240}{x}$

Substituting $x$ in the above equation, we get:

$$\begin{gathered} C=10\times 8^2+\frac{10240}{8} \\ C=640+1280 \\ C=1920 \end{gathered}$$

Hence, the least cost of the box is $₹1920$.