If $x + y = 12$ , then minimum value of $x^{2} + y^{2}$

72

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144

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48

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36

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Solution

The correct option is C $72$
$f (x) = x^{2} + (12 - x)^{2}$
$f^{'} (x) = 2 x + (- 2) (12 - x)$
$0 = 2 (x - 12 + x)$
$x = 6$
$f (6) = 36 + 36$
$= 72$
Now, $f^{''} (x) = 4 > 0$
Thus, minima at $x = 6$
Hence, minimum value is $72$ .

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Similar questions

A

x + y = 12

x^{2} + y^{2}

x + y = k

k^{2}

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) $0$