MHTCET 15 May 2024 Shift 2 Paper

Concept:For an open tank (no top), the surface area is the sum of the square base and four side walls. To minimise this area, we use differentiation and set the first derivative to zero.Explanation:Let $x$ cm be the side of the square bottom and $h$ cm be the height of the tank.The volume is given as:$$V = x^{2}h = 4000$$The surface area of the open tank is:$$A = x^{2} + 4xh$$From the volume equation, express $h$ in terms of $x$:$$h = \frac{4000}{x^{2}}$$Substitute this value of $h$ into the surface area equation:$$A = x^{2} + 4x \cdot \frac{4000}{x^{2}} = x^{2} + \frac{16000}{x}$$Differentiate $A$ with respect to $x$:$$\frac{dA}{dx} = 2x - \frac{16000}{x^{2}}$$For minimum surface area, set $\frac{dA}{dx} = 0$:$$2x - \frac{16000}{x^{2}} = 0$$$$\Rightarrow 2x = \frac{16000}{x^{2}}$$$$\Rightarrow x^{3} = 8000$$$$\Rightarrow x = 20$$Now verify using the second derivative:$$\frac{d^{2}A}{dx^{2}} = 2 + \frac{32000}{x^{3}}$$At $x = 20$:$$\frac{d^{2}A}{dx^{2}} = 2 + \frac{32000}{8000} = 2 + 4 = 6 > 0$$Since the second derivative is positive, $A$ is minimum at $x = 20$ cm.Now find the height:$$h = \frac{4000}{x^{2}} = \frac{4000}{400} = 10 \text{ cm}$$Answer:The dimensions of the tank for minimum surface area are side of square bottom $= 20$ cm and height $= 10$ cm.Therefore, the correct option is B.