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

The correct answer is Option (2) → ₹1,920 ##

Since, volume of the box $= 1024 \text{ cm}^3$

Let length of the side of square base be $x \text{ cm}$ and height of the box be $y \text{ cm}$.

$∴\quad$ Volume of the box $(V) = x^2 \cdot y = 1024$

Since, $\quad x^2 y = 1024 \Rightarrow y = \frac{1024}{x^2} \quad [V = l \times b \times h] \dots (i)$

Let $C$ denotes the cost of the box.

Area of top and bottom $x^2 + x^2 = 2x^2$

Area of the 4 sides $= 4xy$

$∴\quad C = 2x^2 \times 5 + 4xy \times 2.50$

$\quad \quad \quad = 10x^2 + 10xy = 10x(x + y)$

$\quad \quad \quad = 10x \left( x + \frac{1024}{x^2} \right) \quad \text{[using Eq. (i)]}$

$\quad \quad \quad = \frac{10x}{x^2} (x^3 + 1024)$

$\Rightarrow \quad C = 10x^2 + \frac{10240}{x} \quad \dots (ii)$

On differentiating both sides w.r.t. $x$, we get

$\frac{dC}{dx} = 20x + 10240 \left( -\frac{1}{x^2} \right)$

$= 20x - \frac{10240}{x^2} \quad \dots \text{(iii)}$

Now, $\frac{dC}{dx} = 0$

$\Rightarrow 20x = \frac{10240}{x^2}$

$\Rightarrow 20x^3 = 10240$

$\Rightarrow x^3 = 512 = 8^3 \Rightarrow x = 8$

Again, differentiating Eq. (iii) w.r.t. $x$, we get

$\frac{d^2C}{dx^2} = 20 - 10240(-2) \cdot \frac{1}{x^3}$

$= 20 + \frac{20480}{x^3}$

$∴\left( \frac{d^2C}{dx^2} \right)_{x=8} = 20 + \frac{20480}{512} = 60 > 0$

For $x = 8$, cost is minimum and the corresponding least cost of the box,

$C(8) = 10 \cdot 8^2 + \frac{10240}{8}$

$= 640 + 1280 = 1920$

$∴$ Least cost $= ₹ 1920$