MHTCET 26 Apr 2025 Shift 2 Paper

Concept:Use the constraint $xy = c^2$ to rewrite the expression in one variable, then apply the second derivative test to find the minimum.Explanation:Let $p = ax + by$.Since $xy = c^2$, we have $y = \dfrac{c^2}{x}$.Therefore, $p = ax + \dfrac{bc^2}{x}$.Differentiating: $\dfrac{dp}{dx} = a - \dfrac{bc^2}{x^2}$.Differentiating again: $\dfrac{d^2p}{dx^2} = \dfrac{2bc^2}{x^3}$.For stationary points, set $\dfrac{dp}{dx} = 0$: $a - \dfrac{bc^2}{x^2} = 0$.This gives $x^2 = \dfrac{bc^2}{a}$, so $x = \pm c\sqrt{\dfrac{b}{a}}$.At $x = c\sqrt{\dfrac{b}{a}}$, $\dfrac{d^2p}{dx^2} > 0$, so $p$ is minimum at this point.Thus, minimum value is $p_{\min} = a\left(c\sqrt{\dfrac{b}{a}}\right) + \dfrac{bc^2}{c\sqrt{\dfrac{b}{a}}}$.This simplifies to $p_{\min} = c\sqrt{ab} + c\sqrt{ab} = 2c\sqrt{ab}$.Answer:$2c\sqrt{ab}$ — Option A.