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(B) Let the two parts be $x$ and $y$ such that $x + y = 20$. We want to maximize the product $z = x \cdot y^3$. Substituting $x = 20 - y$ into the equ

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$(5, 15)$

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(B) Let the two parts be $x$ and $y$ such that $x + y = 20$. We want to maximize the product $z = x \cdot y^3$. Substituting $x = 20 - y$ into the equation,we get $z = (20 - y)y^3 = 20y^3 - y^4$. To find the critical points,we differentiate $z$ with respect to $y$: $\frac{dz}{dy} = 60y^2 - 4y^3$. Setting $\frac{dz}{dy} = 0$,we get $4y^2(15 - y) = 0$,which gives $y = 0$ or $y = 15$. Now,we find the second derivative: $\frac{d^2z}{dy^2} = 120y - 12y^2$. At $y = 15$,$\frac{d^2z}{dy^2} = 120(15) - 12(15^2) = 1800 - 2700 = -900 Since the second derivative is negative,$y = 15$ is the point of maxima. If $y = 15$,then $x = 20 - 15 = 5$. Thus,the two parts are $(5, 15)$.

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

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