We want to maximize [tex]x^3y^4z[/tex] subject to the constraint [tex]x+y+z=30[/tex].
The Lagrangian is
[tex]L(x,y,z,\lambda)=x^3y^4z-\lambda(x+y+z-30)[/tex]
with critical points where the derivatives vanish:
[tex]L_x=3x^2y^4z-\lambda=0[/tex]
[tex]L_y=4x^3y^3z-\lambda=0[/tex]
[tex]L_z=x^3y^4-\lambda=0[/tex]
[tex]L_\lambda=x+y+z-30=0[/tex]
[tex]\implies\lambda=3x^2y^4z=4x^3y^3z=x^3y^4[/tex]
We have
[tex]3x^2y^4z-4x^3y^3z=x^2y^3z(3y-4x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\z=0,\text{ or}\\3y=4x\end{cases}[/tex]
[tex]3x^2y^4z-x^3y^4=x^2y^4(3z-x)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}\\3z=x\end{cases}[/tex]
[tex]4x^3y^3z-x^3y^4=x^3y^3(4z-y)=0\implies\begin{cases}x=0,\text{ or}\\y=0,\text{ or}4z=y\end{cases}[/tex]
Let's work with [tex]x=3z[/tex] and [tex]y=4z[/tex], for which we have
[tex]x+y+z=8z=30\implies z=\dfrac{15}4\implies\begin{cases}x=\frac{45}4\\y=15\end{cases}[/tex]
The smallest of these is C. 15/4.
