If x and y are two positive numbers such that | vedclass

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(D) Let $s = x^2 + y^2$. Given $x + y = 32$,we can write $y = 32 - x$. Substituting this into the expression for $s$: $s = x^2 + (32 - x)^2$. To find

Vedclass · Accepted Answer

$512$

Vedclass · Answer

(D) Let $s = x^2 + y^2$. Given $x + y = 32$,we can write $y = 32 - x$. Substituting this into the expression for $s$: $s = x^2 + (32 - x)^2$. To find the minimum,we differentiate $s$ with respect to $x$: $\frac{ds}{dx} = 2x + 2(32 - x)(-1) = 2x - 64 + 2x = 4x - 64$. Setting $\frac{ds}{dx} = 0$ for critical points: $4x - 64 = 0 \implies x = 16$. Then $y = 32 - 16 = 16$. Checking the second derivative: $\frac{d^2s}{dx^2} = 4 > 0$,which confirms that $s$ is minimum at $x = 16$. The minimum value is $s = 16^2 + 16^2 = 256 + 256 = 512$.

If $x$ and $y$ are two positive numbers such that $x+y=32$,then the minimum value of $x^2+y^2$ is,

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