Question

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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (A)

Let $x,y$ be two numbers such that
$x+y=3\Rightarrow y=3-x$ and let product $P=x{y}^2$
thus $P=x(3-x{)}^2=x^3-6x^2+9x$
For a maxima or minima $\frac{dP}{dx}=0$
Thus $\frac{dP}{dx}=3x^2-12x+9$ and
$\frac{d^{2}P}{d{x}^2}=6x-12$
Now, $\frac{dP}{dx}=0$
$\Rightarrow 3x^2-12x+9=0$
$\Rightarrow x=1,3$.
Thus ${\left(\frac{d^{2}P}{d{x}^2}\right)}_{x=1}=-6$
and ${\left(\frac{d^{2}P}{d{x}^2}\right)}_{x=3}=6$
Thus $P$ is maximum when $x=1$
$\Rightarrow y=2$ So, $P=1.2^2=4$.