The sum of two numbers is 3, then max value of the product of the first and the square of the second is

4

Let two numbers be x, 3 – x.

∴  $P=x(3-x)^2 \Rightarrow \frac{d P}{d x}=-2 x(3-x)+(3-x)^2$

∴  $\frac{d P}{d x}=(3-x)(3-3 x)$

and  $\frac{d^2 P}{d x^2}=(3-x)(-3)+(3-3 x)(-1)=6 x-12$

$\frac{d P}{d x}=0 \Rightarrow x=3,1$

But x = 3 is not possible

$\left.\frac{d^2 P}{d x^2}\right|_{x=1}=-6<0$

∴ P is max. at x = 1 and maximum value is 1(3 – 1)² = 4.