If $x + y = 3$ , Show that the maximum value of $x^{2} y$ is $4$

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Solution

$x + y = 3$
$y = 3 - x$
$x^{2} y =$
$x^{2} (3 - x)$
$t = 3 x^{2} - x^{3}$
$\frac{d t}{d x} = 6 x - 3 x^{2}$
$= 3 x (2 - x)$
$3 x = 0 ∣ 2 - x = 0$
$x = 0 ∣ x = 2$
Now
$t = (x^{2}) (3 - x)$
$x = 2$
$= (2)^{2} (3 - 2)$
$= 4 \times 1$
$= 4$

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