UPSC NDA II 2023 Mathematics Solved Paper

Concept:The maximum value of a product $xy$ under a given relation is found by maximizing one variable and then using the domain of the other.Explanation:The relation given is $x = \frac{1}{4}\sqrt{9 - y^{2}}$.Differentiate $x$ with respect to $y$ to find its maximum:$\frac{dx}{dy} = \frac{1}{4} \cdot \frac{1}{2\sqrt{9-y^{2}}} \cdot (-2y) = \frac{-y}{4\sqrt{9-y^{2}}}$.Set $\frac{dx}{dy} = 0$ to find critical points: $\frac{-y}{4\sqrt{9-y^{2}}} = 0 \Rightarrow y = 0$.Check second derivative at $y = 0$: $\frac{d^{2}x}{dy^{2}} = -\frac{1}{12} < 0$, so $x$ is maximum at $y = 0$.Maximum value of $x$ is $x_{\text{max}} = \frac{1}{4}\sqrt{9-0^{2}} = \frac{3}{4}$.The domain of $y$ is $-3 \le y \le 3$ (from $\sqrt{9-y^{2}}$), so the maximum possible $y$ is $y_{\text{max}} = 3$.Thus, the maximum value of $xy$ is $x_{\text{max}} \times y_{\text{max}} = \frac{3}{4} \times 3 = \frac{9}{4}$.Answer:$\frac{9}{4}$