If x + y = 16 and x^2 + y^2 is minimum,then t | vedclass

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(D) Given $x + y = 16$,we can express $y$ as $y = 16 - x$.Let $f(x) = x^2 + y^2$. Substituting $y$,we get $f(x) = x^2 + (16 - x)^2$.Expanding this,$f(

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$8, 8$

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(D) Given $x + y = 16$,we can express $y$ as $y = 16 - x$.Let $f(x) = x^2 + y^2$. Substituting $y$,we get $f(x) = x^2 + (16 - x)^2$.Expanding this,$f(x) = x^2 + 256 - 32x + x^2 = 2x^2 - 32x + 256$.To find the minimum,we find the derivative $f'(x) = 4x - 32$.Setting $f'(x) = 0$,we get $4x = 32$,which implies $x = 8$.Since $f''(x) = 4 > 0$,the function has a minimum at $x = 8$.Substituting $x = 8$ into $y = 16 - x$,we get $y = 16 - 8 = 8$.Thus,the values are $x = 8$ and $y = 8$.

If $x + y = 16$ and $x^2 + y^2$ is minimum,then the values of $x$ and $y$ are

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