# Maximum/minimum : $u = xy^2z^3$

Find the maximum and minimum values (if they exists) of $u=xy^2z^3$ when $x+y+z = 12$ ($x>0, y>0, z>0$)

It is not hard to find the possible points of maximum/minimum of this function using Lagrange multipliers and we can see that $x,y,z \leq 12$. But, how can I be sure about the existance (or not) of maximum/minimum of this function?

Thanks!

• If you allow $x,y,z\geq 0$ the region is compact, so there must be a maximum. But none of tha cases where $x=0,y=0.$ or $z=0$ can lead to a maximum. Nov 13 '13 at 13:11
• Indeed, since $x,y,z > 0$. But how about minimum? Nov 13 '13 at 13:22
• No absolute minimum exists since we can get as close to zero as possible but can't reach zero. Nov 13 '13 at 13:26
• Ok, thank you very much. Nov 13 '13 at 13:28

For maximum, notice that $12=x+y+z=x+2\frac y2+3\frac z3$, then apply AM-GM inequality $$xy^2z^3=108(x)(\frac y2)^2(\frac z3)^3\leq108\left(\frac{x+2\frac y2+3\frac z3}6\right)^6=6912$$ "=" holds iff $x=\frac y2=\frac z3=2$, or $x=2,y=4,z=6$.

For minimum part, let $x$ tends to zero, then $xy^2z^3$ also tends to zero. So there's no minimum.

You can trace the domain of possible values possible in a 3D plan.

1) Find an expression for x $$x = 12 - y - z$$ 2) Substitute x in the first equation $$u = (12 - y - z)y^{2}z^{3} \\ u = 12y^{2}z^{3} - y^{3}z^{3} - z^{4}$$ 3) Draw a 3D graph using the last equation while varying the z and y parameters using these constraints: $$y>0,z>0$$

4) When finding the maximum, you can find the x coordinate with the first equation. I hope this works for you.

Just by looking at the equation of the graph, it is sure there is a maximum because every limit tend to infinity:

$$y=0,\lim_{z \to +\infty} -z^{4}= -\infty \\ z=0,\lim_{y \to +\infty} 0 = 0 \\$$

Therefore, with the superposition theorem, the limit for z and y tending to infinity is in the interval $$[0,-\infty]$$. If we can find a single set of values for z and y that make u greater than 0, a maximum exists i.e. $$(y,z) = (1,1) \\ u(1,1) = 10$$ By finding a single set of value that is superior to the interval found, we proved there exists a maximum.

Can express $x$ as $$x = 12 - y - z$$

Using this $u$ becomes

$$u(y,z) = 12y^2z^3 -y^3z^3 - y^2z^4$$

Partial derivatives of $u(y,z)$:

$$\partial_y u = 24yz^3 - 3y^2z^3 -2yz^4$$

and

$$\partial_z u =36y^2z^2 - 3y^3z^2 - 4y^2z^3$$

At an extremum points, the partial derivatives will be zero so:

$$24yz^3 - 3y^2z^3 -2yz^4 = 0$$ divide by $yz^3$ $$24-3y - 2z =0$$

and

$$36y^2z^2 - 3y^3z^2 - 4y^2z^3 = 0$$ divide by $y^2z^2$ $$36 - 3y - 4z = 0$$

From these we get $y = 4$ and $z = 6$. From the original condition on $x,y,z$ which is that $x+y+z=12$ we get $x = 2$.

This is the only solution in this case (where neither $y$ or $z$ are $0$) so assume this gives maximum which is

$$u(2,4,6) = (2)(16)(216)$$

One can check if this point is really a minimum, maximum or saddle point by using second derivative test (http://mathworld.wolfram.com/SecondDerivativeTest.html).