The maximum value of $x y$ subject to $x + y = 4$ is

8

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Solution

The correct option is B $4$
$x + y = 4$
Assume, $A = x y$
$A = x (4 - x)$
$A = 4 x - x^{2}$
For A to be maximum $\frac{d A}{d x} = 0 and \frac{d^{2} A}{d x^{2}} < 0$
$\frac{d A}{d x} = 4 - 2 x = 0$
$∴ x = 2$
$\frac{d^{2} A}{d x^{2}} = - 2 < 0$
Hence the maximum value of $A = x y = 4$

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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) $0$

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