If $x y (y - x) = 2 a^{3}$ , at what point does y have a minimum value .

a, 2a

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a,-a

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2a,2a

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None of these

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Solution

The correct option is D a, 2a
$x y (y - x) = 2 a^{3}$
$x y^{2} - x^{2} y = 2 a^{3}$
Differentiating with respect to x, we get
$y^{2} + 2 x y . y^{'} - 2 x y - x^{2} y^{'} = 0$
Or
$y^{2} - 2 x y + y^{'} (2 x y - x^{2}) = 0$
Or
$y^{'} = \frac{2 x y - y^{2}}{2 x y - x^{2}}$
$= 0$
Hence
$2 x y - y^{2} = 0$
$y (2 x - y) = 0$
$y = 0$ and $y = 2 x$
Substituting $x = \frac{y}{2}$ , we get
$\frac{y^{2}}{2} (\frac{y}{2}) = 2 a^{3}$
$y^{3} = 8 a^{3}$
Or
$y = 2 a$
Hence
$x = a$ .

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