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