Given $f(x,y,z)=xy^2z$ find the the max value such that the point $(x,y,z)$ is located in the part of the plane $x+y+z=4$

Question

Given $f(x,y,z)=xy^2z$ find the the max value such that the point $(x,y,z)$ is located in the part of the plane $x+y+z=4$ which is in the first octant ($x>0, y>0, z>0$) of the coordinates.

I think Lagrange multipliers can be used here.

We can say that $x+y+z=4$ is the restriction. Let $g=x+y+z-4$. Then: $$ \nabla f= \lambda\nabla g\\ \nabla f=\langle y^2z,2yxz,xy^2 \rangle\\ \nabla g=\langle 1,1,1 \rangle $$ It follows that: $$ \begin{cases} y^2z=\lambda\\ 2yxz=\lambda\\ xy^2=\lambda \end{cases} $$ Because $x>0, y>0, z>0$ we can solve the system and get: $$ z=x,\quad y=2x $$ Finally, we can rewrite the restriction as: $$ x+2x+x=4 \Rightarrow x=1, y=2, z=1 $$

Is my usage of Lagrange method correct? Can this problem also be solved without Lagrange method by finding the max value using min/max theorems for functions with 2 variables?

Answer 1

Your method is correct, it might be worth checking this is a maximum though. You could solve it by writing $z = 4 - x - y$, and writing $f$ in terms of $x$ and $y$ only, then minimising with respect to them, but I'd personally go with Lagrange multipliers first.

Answer 2

Can this problem also be solved without Lagrange method by finding the max value using min/max theorems for functions with 2 variables?

Yes; since this constraint surface is relatively simple, we can in fact eliminate one variable and solve directly in terms of $x$ and $y$. On the plane we have $z = 4 - x - y$, and so the restriction of $f$ to the given plane is $$ \bar{f}(x,y) = f(x,y,4-x-y) = xy^2(4-x-y) = 4 xy^2 - x^2 y^2 - x y^3. $$ Taking the gradient of $\bar{f}$ with respect to $x$ and $y$ we obtain $$ \vec{\nabla} \bar{f} = (4y^2 - 2xy^2 - y^3) \hat{\imath} + (8xy - 2x^2y - 3xy^2) \hat{\jmath} = y^2 (4 - 2x - y) \hat{\imath} + xy (8 - 2x - 3y) \hat{\jmath} $$ Since we are restricting to $x > 0$ and $y > 0$, this implies that the extremum occurs at $$ \begin{align*} 2x + y - 4 &= 0 \\ 2x + 3y - 8 &= 0 \end{align*} $$ which can easily be solved to find that $x = 1$, $y = 2$ as you found above.

Answer 3

By letting $y=4-x-z$, we consider the function $g(x,z)=xz(4-x-z)^2$ in the compact set given by the triangle $T$ of vertices $(0,0)$, $(0,4)$ and $(4,0)$. Note that the continuous (and differentiable) function $g$ is zero on the boundary of $T$ and it is positive inside. So, by Weierstrass theorem, the maximum value of $g$ is attained at some critical point inside $T$. By computing the partial derivative $g_x$ and $g_y$ it is easy to see that there is just one critical point inside $T$: if $(x,z)$ is inside $T$ then $$(g_x,g_y)=(z (x+z-4)(3x + z-4),x (x+z-4)(x+3z-4))=(0,0)\implies (x,z)=(1,1).$$ Therefore $(1,4-1-1,1)=(1,2,1)$ is a maximum point (the same of yours).

Answer 4

By AM-GM $$xy^2z=\frac{1}{4}\left(2x\cdot y\cdot y\cdot2z\right)\leq\frac{1}{4}\left(\frac{2x+y+y+2z}{4}\right)^4=4.$$ The equality occurs for $x=z=1$ and $y=2$, which gives the answer: $4$.

Done!