Maximize [tex]f(x,y,z)=8xyz^2-200(x+y+z)[/tex] subject to [tex]x+y+z=100[/tex]. The Lagrangian is
[tex]L(x,y,z,\lambda)=8xyz^2-200(x+y+z)+\lambda(x+y+z-100)[/tex]
with partial derivatives (set equal to 0):
[tex]L_x=8yz^2-200+\lambda=0[/tex]
[tex]L_y=8xz^2-200+\lambda=0[/tex]
[tex]L_z=16xyz-200+\lambda=0[/tex]
[tex]L_\lambda=x+y+z-100=0[/tex]
From the first two equations it follows that [tex]x=y[/tex]. Subtracting either the first or second equation from the third tells us that
[tex]16xyz-8xz^2=8xz(2y-z)=0[/tex]
which means either [tex]x=0[/tex], [tex]z=0[/tex], or [tex]2y=2x=z[/tex].
If [tex]x=0[/tex], then [tex]y=0[/tex], so that [tex]0+0+z=100[/tex]. So one critical point is (0, 0, 100).
If [tex]z=0[/tex], then [tex]x+y+0=2x=100\implies x=y=50[/tex]. So another critical point is (50, 50, 0).
If [tex]2x=2y=z[/tex], then [tex]x+y+z=x+x+2x=100\implies x=y=25\implies z=50[/tex]. So our third critical point is (25, 25, 50).
Evaluate the revenue function for each of these three critical points. You'll find that the latter critical point achieves the maximum value of 12,480,000 in revenue.
