Answer to Question #287929 in Differential Equations for Varun

Question

Find the maximum value of

x^2y^3z^4

subject to the condition

x+y+z=5

Answer 1

$\text{Let, } a=x^2y^3z^4 \Rightarrow \text{ln } a=2\text{ ln } x+3\text{ ln } y+4\text{ ln } z\\$

Let

$P=ln (a)+\lambda (x+y+z-5)\\ \therefore \frac{\partial P}{\partial x}=\frac{2}{x}+\lambda,~ \frac{\partial P}{\partial y}=\frac{3}{y}+\lambda, ~ \frac{\partial P}{\partial z}=\frac{4}{z}+\lambda$

If we maximizing a implies maximizing P.

i.e. $\frac{\partial P}{\partial x}=\frac{\partial P}{\partial y}=\frac{\partial P}{\partial z}=0$

$\Rightarrow \lambda=-\frac{2}{x}=-\frac{3}{y}=-\frac{4}{z}\\ \therefore \frac{2}{x}=\frac{3}{y}=\frac{4}{z}= \frac{2+3+4}{x+y+z}=\frac{9}{5} \\ \texttt{[As we are doing this under the constraint x+y+z=5]}\\$

$\Rightarrow x=\frac{10}{9}, y=\frac{15}{9}, z=\frac{20}{9}\\ \text{Hence the maximum value of $a$ is }\\ \max_{x,y,z}a=(\frac{10}{9})^2\times(\frac{5}{3})^3\times(\frac{10}{9})^4=\dfrac{2\times10^9}{3^{15}}$