Question

Let $P ( x )=\frac{5}{3}-6 x -9 x ^{2}$ and $Q ( y )=-4 y ^{2}+4 y +\frac{13}{2}$ If there exists unique pair of real number $(x, y)$ such that $P ( x ) Q ( y )=20$, then the value of $(6 x +10 y )$ is _____

Answer 1

We have
$P ( x )=\frac{5}{3}-6 x -9 x ^{2}=-(3 x +1)^{2}+\frac{8}{3}$
$ \Rightarrow P _{\max }=\frac{8}{3}$
Similarly, $Q(y)=-4 y^{2}+4 y+\frac{13}{2}$
$=-(2 y-1)^{2}+\frac{15}{2}$
$\Rightarrow Q_{\max }=\frac{15}{2}$
Now, $P_{\max } \times Q_{\max }=\frac{8}{3} \times \frac{15}{2}=20$
So $( x , y ) \equiv\left(-\frac{1}{3}, \frac{1}{2}\right)$
Hence, $6 x+10 y=6\left(\frac{-1}{3}\right)+10\left(\frac{1}{2}\right)$
$=-2+5=3$