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(D) Let the two parts be $x$ and $y$ such that $x + y = 20$,which implies $y = 20 - x$.We want to maximize the product $P = x^3 y^2$.Substituting $y$,

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$12, 8$

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(D) Let the two parts be $x$ and $y$ such that $x + y = 20$,which implies $y = 20 - x$.We want to maximize the product $P = x^3 y^2$.Substituting $y$,we get $P(x) = x^3 (20 - x)^2 = x^3 (400 - 40x + x^2) = 400x^3 - 40x^4 + x^5$.To find the critical points,differentiate $P$ with respect to $x$:$\frac{dP}{dx} = 1200x^2 - 160x^3 + 5x^4$.Setting $\frac{dP}{dx} = 0$:$5x^2 (240 - 32x + x^2) = 0$.$5x^2 (x - 12)(x - 20) = 0$.Thus,$x = 0, 12, 20$. Since the parts must be positive,we consider $x = 12$.Using the second derivative test: $\frac{d^2P}{dx^2} = 2400x - 480x^2 + 20x^3$.At $x = 12$,$\frac{d^2P}{dx^2} = 2400(12) - 480(144) + 20(1728) = 28800 - 69120 + 34560 = -5760 Since the second derivative is negative,$x = 12$ is a point of local maxima.Therefore,the parts are $x = 12$ and $y = 20 - 12 = 8$.

$20$ is divided into two parts such that the product of the cube of one part and the square of the other part is maximum. The parts are:

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