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 $y = \frac{8 - x}{2}$.Let the function to be maximized be $f(x) = xy$. Sub

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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 $y = \frac{8 - x}{2}$.Let the function to be maximized be $f(x) = xy$. Substituting the expression for $y$,we get $f(x) = x \left( \frac{8 - x}{2} ight) = 4x - \frac{x^2}{2}$.To find the critical points,we calculate the first derivative: $f'(x) = \frac{d}{dx} (4x - \frac{x^2}{2}) = 4 - x$.Setting $f'(x) = 0$,we get $4 - x = 0$,which implies $x = 4$.Substituting $x = 4$ into the constraint equation,we find $y = \frac{8 - 4}{2} = 2$.To verify that this is a maximum,we check the second derivative: $f''(x) = -1$. Since $f''(x) The maximum value is $f(4) = 4 	imes 2 = 8$.

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

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