If $2\mathrm{p}+3\mathrm{q}+4\mathrm{r}=15$ then the maximum  value of  ${\mathrm{p}}^3{\mathrm{q}}^5{\mathrm{r}}^7$ is 

Since,$\frac{\frac{2\mathrm{p}}{3}+\frac{2\mathrm{p}}{3}+\frac{2\mathrm{p}}{3}+\underset{5\text{ times }}{\underbrace{\frac{3\mathrm{q}}{5}+\dots +\frac{3\mathrm{q}}{5}}}+\underset{7\text{ times }}{\underbrace{\frac{4\mathrm{r}}{7}+\dots +\frac{4\mathrm{r}}{7}}}}{15}$

  $\ge \sqrt[15]{{\left(\frac{2\mathrm{p}}{3}\right)}^3{\left(\frac{3\mathrm{q}}{5}\right)}^5{\left(\frac{4\mathrm{r}}{7}\right)}^7}$   [AM$\ge$GM]

$\Rightarrow {\mathrm{p}}^3{\mathrm{q}}^5{\mathrm{r}}^7\frac{2^3{3}^5{4}^7}{3^3{5}^5{7}^7}\le 1\Rightarrow {\mathrm{p}}^3{\mathrm{q}}^5{\mathrm{r}}^7\le \frac{5^5{7}^7}{2^3{3}^2{4}^7}\le \frac{5^5{7}^7}{2^{17}9}$