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

Let the two numbers be a and b .
\(\begin{array}{ll}
\therefore & a+b=3 \\
\therefore & b=3-a...(i)
\end{array}\)
\(\therefore \quad\) Product of first number and square of second number \((p)=a b^2\)
\(\mathrm{p}=\mathrm{a}(3-\mathrm{a})^2\)
\(\ldots[\) From (i)]
\(\begin{array}{ll}
\therefore & p=9 a-6 a^2+a^3 ...(ii)\\
\therefore & \frac{d p}{d a}=9-12 a+3 a^2 \\
\therefore & \frac{d^2 p}{d a^2}=-12+6 a
\end{array}\)
Now, \(\frac{\mathrm{dp}}{\mathrm{da}}=0 \Rightarrow 3 \mathrm{a}^2-12 \mathrm{a}+9=0\)
\(\begin{aligned}
& \Rightarrow(3 a-9)(a-1)=0 \\
& \Rightarrow a=3 \text { or } a=1
\end{aligned}\)
\(\left.\frac{\mathrm{d}^2 p}{d \mathrm{da}^2}\right|_{a=3}=6\gt0 \text { and }\left.\frac{\mathrm{d}^2 p}{\mathrm{da}^2}\right|_{\mathrm{a}=1}=-6 \lt 0\)
\(\therefore \quad\) Maximum value of p is at \(\mathrm{a}=1\)
\(\therefore \quad\) Maximum value of the product \(=4\)
...[From (ii)]