If x + 2y = 8,then the maximum value of xy is | vedclass

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(D) Given the equation $x + 2y = 8$,we can express $x$ as $x = 8 - 2y$.Let $f(y) = xy = (8 - 2y)y = 8y - 2y^2$.To find the maximum value,we differenti

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$8$

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(D) Given the equation $x + 2y = 8$,we can express $x$ as $x = 8 - 2y$.Let $f(y) = xy = (8 - 2y)y = 8y - 2y^2$.To find the maximum value,we differentiate $f(y)$ with respect to $y$:$f'(y) = \frac{d}{dy}(8y - 2y^2) = 8 - 4y$.Setting the first derivative to zero for critical points:$8 - 4y = 0 \implies y = 2$.Now,we find the second derivative to check for maxima:$f''(y) = -4$.Since $f''(2) = -4 Substituting $y = 2$ back into the expression for $f(y)$:$f(2) = 8(2) - 2(2)^2 = 16 - 8 = 8$.Thus,the maximum value of $xy$ is $8$.

If $x + 2y = 8$,then the maximum value of $xy$ is .......

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