The sum of two numbers is 20. If the product | vedclass

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Question

(A) Let the two numbers be $x$ and $y$. Given that $x + y = 20$,so $y = 20 - x$. Let the function to be maximized be $f(x) = x^2 y^3 = x^2 (20 - x)^3$

Vedclass · Accepted Answer

$12, 8$

Vedclass · Answer

(A) Let the two numbers be $x$ and $y$. Given that $x + y = 20$,so $y = 20 - x$. Let the function to be maximized be $f(x) = x^2 y^3 = x^2 (20 - x)^3$. To find the maximum,differentiate $f(x)$ with respect to $x$: $f'(x) = 2x(20 - x)^3 + x^2 \cdot 3(20 - x)^2 (-1)$ $f'(x) = x(20 - x)^2 [2(20 - x) - 3x]$ $f'(x) = x(20 - x)^2 [40 - 2x - 3x] = x(20 - x)^2 (40 - 5x)$ Setting $f'(x) = 0$,we get $x = 0$,$x = 20$,or $5x = 40 \Rightarrow x = 8$. Since $x$ and $y$ must be positive,we test $x = 8$. If $x = 8$,then $y = 20 - 8 = 12$. Thus,the numbers are $8$ and $12$.

The sum of two numbers is $20$. If the product of the square of one number and the cube of the other is maximum,then the numbers are:

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