If x and y are two positive integers such tha | vedclass

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(C) Given $x+y=24$,so $y=24-x$. Let $P = x^3 y^5 = x^3(24-x)^5$. To find the maximum,we differentiate $P$ with respect to $x$: $\frac{dP}{dx} = 3x^2(2

Vedclass · Accepted Answer

$306$

Vedclass · Answer

(C) Given $x+y=24$,so $y=24-x$. Let $P = x^3 y^5 = x^3(24-x)^5$. To find the maximum,we differentiate $P$ with respect to $x$: $\frac{dP}{dx} = 3x^2(24-x)^5 + x^3 \cdot 5(24-x)^4(-1) = 0$. $x^2(24-x)^4 [3(24-x) - 5x] = 0$. $x^2(24-x)^4 [72 - 3x - 5x] = 0$. $x^2(24-x)^4 (72 - 8x) = 0$. Since $x$ and $y$ are positive integers,$x 
eq 0$ and $x 
eq 24$. Thus,$72 - 8x = 0 \Rightarrow x = 9$. Then $y = 24 - 9 = 15$. Finally,$x^2 + y^2 = 9^2 + 15^2 = 81 + 225 = 306$.

If $x$ and $y$ are two positive integers such that $x+y=24$ and $x^3 y^5$ is maximum,then $x^2+y^2=$

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