The maximum value of xy when x+2y=8 is | vedclass

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(D) Given the constraint $x+2y=8$. We can express $y$ in terms of $x$ as $2y = 8-x$,which gives $y = \frac{8-x}{2}$. Let the function to be maximized

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$8$

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(D) Given the constraint $x+2y=8$. We can express $y$ in terms of $x$ as $2y = 8-x$,which gives $y = \frac{8-x}{2}$. Let the function to be maximized be $f(x) = xy$. Substituting the value of $y$,we get $f(x) = x \cdot \frac{8-x}{2} = 4x - \frac{x^2}{2}$. To find the maximum value,we differentiate $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}(4x - \frac{x^2}{2}) = 4 - x$. Setting the derivative to zero for critical points: $4 - x = 0 \implies x = 4$. To confirm it is a maximum,we check the second derivative: $f''(x) = -1$. Since $f''(4) = -1 Substituting $x = 4$ back into the expression for $y$: $y = \frac{8-4}{2} = 2$. The maximum value of $xy$ is $4 	imes 2 = 8$.

The maximum value of $xy$ when $x+2y=8$ is

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