The maximum value of xy subject to x + y = 7 is

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\frac{49}{4}

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\frac{55}{4}

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Solution

The correct option is A $\frac{49}{4}$
$f (x) = x (7 - x)$
$f^{'} (x) = 7 - x - x$
$x = \frac{7}{2}$ at $f^{'} (x) = 0$
$y = \frac{7}{2}$
xy is max at $x = \frac{7}{2}$ $y = \frac{7}{2}$
$= \frac{49}{4}$

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Similar questions

x y

x + y = 4

List-I	List-II
(A) Maximum value of $x y$ subject to $x + y = 7$ is	1) $72$
(B) If $l^{2} + m^{2} = 1$ , then the maximum value of $l + m$ is	2) $1$
(C) If $x + y = 12$ , then the minimum Value of $x^{2} + y^{2}$ is	3) $\sqrt{2}$
(D) Minimum value $x^{2} - 8 x + 17$ is	4) $\frac{49}{4}$
	5) $0$

Find the maximum value of z = 3x + 4y subject to constraints x + y ≤ 4, x ≥ 0 and y ≥ 0

y = | x - 4 | - | x - 7 |

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