geometric program maximizing using Arithmetic-Geometric mean inequality

Question

Maximize $xy^2z^3$ subject to $x^3+y^2+z = 39$ and $x,y,z > 0$.

I have that $39 = x^3 + y^2 + z = ..$

I am unsure what value I should use for $\delta_i$ in each coefficient when using the A-G inequality.

Any tips would be greatly appreciated!

Answer 1

Hint

$$39=x^3+3(y^2/3)+9(z/9)\ge 13\sqrt[13]{x^3y^6z^9/3^{21}}$$

Answer 2

Have you tried setting up a Lagrangian? The problem can be written as $$\text{max} \ xy^2z^3+\lambda(39-x^3-y^2-z)$$ Taking first order conditions, you get that $$df/dx=y^2z^2-3x^2\lambda=0$$ $$df/dy=2xyz^3-2y\lambda=0$$ $$df/dz=3xy^2z^2-\lambda=0$$ From here, you can see that $$\lambda=3xy^2z^2$$ Plugging this into the two other equations, you get $$y^2z^3-3x^2(3xy^2z^2)=0$$ $$2xyz^3-2y(3xy^2z^2)=0$$ Getting rid of redundant variables, you get the following system of equations $$z=9x^3$$ $$3y^2=z$$ $$39=x^3+y^2+z$$ You can easily solve this system of equations to get the optimal values of $x,y,z$.