A wire of length 20 units is divided into two | vedclass

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(B) Let the two parts be $x$ and $y$. Given $x + y = 20$,so $y = 20 - x$.Let the function be $f(x) = x(20 - x)^3$.To find the maximum,we differentiate

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) Let the two parts be $x$ and $y$. Given $x + y = 20$,so $y = 20 - x$.Let the function be $f(x) = x(20 - x)^3$.To find the maximum,we differentiate $f(x)$ with respect to $x$:$f'(x) = (20 - x)^3 + x \cdot 3(20 - x)^2(-1)$$f'(x) = (20 - x)^2 [20 - x - 3x] = (20 - x)^2 (20 - 4x)$.Setting $f'(x) = 0$,we get $x = 20$ or $x = 5$.Since $x=20$ gives $f(x)=0$ (minimum),we check $x = 5$.For $x = 5$,$y = 20 - 5 = 15$.The product of the two parts is $xy = 5 	imes 15 = 75$.

$A$ wire of length $20$ units is divided into two parts such that the product of one part and the cube of the other part is maximum. Then,the product of these parts is:

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