Question

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

Answer 1

Answer: 3

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