Question

The sum of two numbers is $z$, the maximum value of the product of the first and the square of second is

• A. $4$
• B. $1$
• C. $3$
• D. $0$

Answer 1

The correct option is A $4$

Let the numbers are $x$ and $(3-x)$
Thus their product is $P=x(3-x{)}^2$
$\Rightarrow \frac{dP}{dx}=-2x(3-x)+(3-x{)}^2$
$\Rightarrow \frac{dP}{dx}=(3-x)(3-3x)$
and $\frac{d^{2}P}{d{x}^2}=6x-12$
For maxima or minima, $\frac{dP}{dx}=0$
$\Rightarrow (3-x)(3-3x)=0$
$\Rightarrow x=3,1$
At $(x=3)$, we have
$\frac{d^{2}P}{d{x}^2}=18-12=6>0$ (minima)
At $(x=1),\frac{d^{2}P}{d{x}^2}=-6<0$
So, $P$ is maximum at $x=1$
Maximum value of $P=1(3-1{)}^2=4$.