Let $\mathrm{P}\left(\mathrm{x}\right)=\frac{3}{3}-6\mathrm{x}-9{\mathrm{x}}^2\text{ and }\mathrm{Q}\left(\mathrm{y}\right)=-4{\mathrm{y}}^2+4\mathrm{y}+\frac{13}{2}$If there exists unique pair of real numbers (x, y) such that $\mathrm{P}\left(\mathrm{x}\right)\cdot \mathrm{Q}\left(\mathrm{y}\right)=20$. then the value of $\frac{1}{\mathrm{xy}}$ is____.

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