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If $x y (x - y) = 2 a^{3} (a > 0)$ then y has minimum when $x =$

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$\frac{1}{a}$

-a

$\frac{a}{2}$

answer is B.

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Detailed Solution

$\frac{d y}{d x} = \frac{- (2 x y - y^{2})}{(x^{2} - 2 x y)} = 0 \Rightarrow y = 2 x$

$\Rightarrow x (2 x) (x - 2 x) = 2 a \Rightarrow - x^{3} = a^{3} \Rightarrow x = - a$

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