If x+y=k, x0, y0,then x^2+y^2 is minimum,if | vedclass

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(C) Given $x+y=k$,where $x>0$ and $y>0$.We can write $y=k-x$.Let $f(x) = x^2+y^2 = x^2+(k-x)^2$.Expanding the expression: $f(x) = x^2+k^2-2kx+x^2 = 2x

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$x=y$

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(C) Given $x+y=k$,where $x>0$ and $y>0$.We can write $y=k-x$.Let $f(x) = x^2+y^2 = x^2+(k-x)^2$.Expanding the expression: $f(x) = x^2+k^2-2kx+x^2 = 2x^2-2kx+k^2$.To find the minimum,we differentiate $f(x)$ with respect to $x$ and set it to zero:$f'(x) = 4x-2k = 0 \Rightarrow x = \frac{k}{2}$.Now,check the second derivative:$f''(x) = 4 > 0$.Since the second derivative is positive,the function has a minimum at $x = \frac{k}{2}$.Substituting $x = \frac{k}{2}$ into $y = k-x$,we get $y = k - \frac{k}{2} = \frac{k}{2}$.Thus,$x=y=\frac{k}{2}$ gives the minimum value of $x^2+y^2$.

If $x+y=k, x>0, y>0$,then $x^2+y^2$ is minimum,if

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