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(D) Given $2x + 3y = 50$. We want to maximize $P = x^2 y^3$. From the constraint, $y = \frac{50 - 2x}{3}$. Substituting $y$ in $P$, we get $P = x^2 \l

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$7$

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(D) Given $2x + 3y = 50$. We want to maximize $P = x^2 y^3$. From the constraint, $y = \frac{50 - 2x}{3}$. Substituting $y$ in $P$, we get $P = x^2 \left(\frac{50 - 2x}{3}ight)^3 = \frac{1}{27} x^2 (50 - 2x)^3$. To find the maximum, we differentiate $P$ with respect to $x$: $\frac{dP}{dx} = \frac{1}{27} [x^2 \cdot 3(50 - 2x)^2(-2) + 2x(50 - 2x)^3] = 0$. $\Rightarrow 2x(50 - 2x)^2 [-3x + 50 - 2x] = 0$. $\Rightarrow 2x(50 - 2x)^2 (50 - 5x) = 0$. Since $x, y$ are positive integers, $x 
eq 0$ and $x 
eq 25$. Thus, $50 - 5x = 0 \Rightarrow x = 10$. For $x = 10$, $y = \frac{50 - 2(10)}{3} = \frac{30}{3} = 10$. Thus, $\alpha = 10$ and $\beta = 10$. Finally, $\frac{\alpha}{2} + \frac{\beta}{5} = \frac{10}{2} + \frac{10}{5} = 5 + 2 = 7$.

$x$ and $y$ are two positive integers such that $2x + 3y = 50$. If $x^2 y^3$ is maximum for $x = \alpha$ and $y = \beta$, then $\frac{\alpha}{2} + \frac{\beta}{5} =$

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