Question

If sum of two numbers is $3$, then maximum value of the product of first and the square of second is

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is A

$4$

Explanation for the correct option:

Step1: Suppose that the first number is $3-x$and let the second number be $x$

$\begin{array}{c}f\left(x\right)=\left(3-x\right)x^2\\ =3x^2-x^3\end{array}$

Step2: Determine the first and second derivatives.

$\begin{array}{c}f\text{'}\left(x\right)=0\\ x=0,2\\ \mathrm{Now},f\text{'}\text{'}\left(x\right)=6-6x\\ \therefore f\text{'}\text{'}\left(2\right)=-6<0\end{array}$

Maximum value is $\left(3-2\right)\times 2^2=4$

Hence option (1) is the answer