If 20 is divided into two parts such that the | vedclass

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(C) Let the two parts be $x$ and $20-x$. Let the product be $P(x) = x^3(20-x)^2$. To find the maximum,we differentiate $P(x)$ with respect to $x$: $P'

Vedclass · Accepted Answer

$12, 8$

Vedclass · Answer

(C) Let the two parts be $x$ and $20-x$. Let the product be $P(x) = x^3(20-x)^2$. To find the maximum,we differentiate $P(x)$ with respect to $x$: $P'(x) = 3x^2(20-x)^2 + x^3 \cdot 2(20-x)(-1)$ $P'(x) = x^2(20-x) [3(20-x) - 2x]$ $P'(x) = x^2(20-x) [60 - 3x - 2x] = x^2(20-x)(60-5x)$. Setting $P'(x) = 0$,we get $x=0$,$x=20$,or $x=12$. Since $x$ must be between $0$ and $20$,we test $x=12$. For $x=12$,the parts are $12$ and $20-12=8$. Thus,the two parts are $12$ and $8$.

If $20$ is divided into two parts such that the product of the cube of one part and the square of the other part is maximum,then these two parts are:

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