# maximum/minimum in multivariable functions

#### mikoamar

##### New member
suppose i have a function f(x,y,z) and i am looking for the minimum value of the function under some restrictions,
using Lagrange multipliers, and i ended up with only 1 suspected point, how can i know if the point is the minimum or maximum?

example:
f(x,y,z)=600xy+900xz+900yz
restriction: g(x,y,z)=xyz-486
i found that the suspected point is M(9,9,6)

#### tkhunny

##### Moderator
Staff member
Are you sure those are your only choices?
How far is it?
How far is (9.1,9,6)?
Maybe (9,9,6.1)?
Modern calculating devices make this test rather simple.

#### MarkFL

##### Super Moderator
Staff member
Normally, the constraint is an equation...should it be:

$$\displaystyle g(x,y,z)=xyz-486=0$$ ?

#### mikoamar

##### New member
Normally, the constraint is an equation...should it be:

$$\displaystyle g(x,y,z)=xyz-486=0$$ ?
yes

#### MarkFL

##### Super Moderator
Staff member
Okay, so then we get the system:

$$\displaystyle 600y+900z=\lambda(yz)$$

$$\displaystyle 600x+900z=\lambda(xz)$$

$$\displaystyle 900x+900y=\lambda(xy)$$

The first two equations imply $$x=y$$, and thus:

$$\displaystyle \frac{600x+900z}{xz}=\frac{1800}{x}$$

$$\displaystyle z=\frac{2}{3}x$$

Substituting for $$y$$ and $$z$$ into the constraint, we obtain:

$$\displaystyle x\cdot x\cdot\frac{2}{3}x=486\implies x=9$$

Thus, our critical point is:

$$\displaystyle (x,y,z)=(9,9,6)\quad\checkmark$$

Okay, we find our objective function at this point is:

$$\displaystyle f(9,9,6)=145800$$

Pick another point on the constraint, observing that $$486=2\cdot3^5$$, and evaluate the objective function at that point to ensure it is greater than $$f(9,9,6)$$ to allow us to conclude the critical point is at a minimum.

#### JeffM

##### Elite Member
$$\displaystyle 600y+900z=\lambda(yz)$$

$$\displaystyle 600x+900z=\lambda(xz)$$

$$\displaystyle 900x+900y=\lambda(xy)$$

The first two equations imply $$x=y$$
At least to me, this was far from obvious.

$$\displaystyle 600y + 900z = \lambda yz \implies \lambda yz - 600y = 900z \implies y( \lambda z - 600) = 900z.$$

$$\displaystyle \therefore \lambda z \ne 600 \implies y = \dfrac{900z}{ \lambda z - 600}.$$

$$\displaystyle 600x + 900z = \lambda xz \implies \lambda xz - 600x = 900z \implies x(\lambda z - 600) = 900z.$$

$$\displaystyle \therefore z \ne 600 \implies x = \dfrac{900z}{\lambda z - 600}.$$

$$\displaystyle \therefore \lambda z \ne 600 \implies x = y.$$

But what if $$\displaystyle \lambda z = 600.$$

$$\displaystyle y( \lambda z - 600) = 900z \implies y * 0 = 900z \implies$$

$$\displaystyle z = 0 \implies xyz \ne 486.$$

So the constraint prevents $$\displaystyle \lambda z = 600.$$

Brilliant. Only took me an hour to figure it out.

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#### MarkFL

##### Super Moderator
Staff member
Interestingly, W|A finds another local minimum, which is less than the one we've found here, but I don't see how to get there using LM:

#### JeffM

##### Elite Member
Interestingly, W|A finds another local minimum, which is less than the one we've found here, but I don't see how to get there using LM:

That possibility worried me as well, but I got sidetracked almost immediately by trying to figure out your simple answer. I suspect that we must take the entire system, which involves four non-linear equations in four unknowns, to get there. It makes my head ache to think about it.

#### JeffM

##### Elite Member
This problem has annoyed me all day.

$$\displaystyle \text {Find extrema of } axy + bxz + byz \text { subject to } xyz = c,\ a \ne 0,\ b \ne 0, \ c \ne 0.$$

$$\displaystyle \therefore x \ne 0,\ y \ne 0, \text { and } z \ne 0.$$

Using Lagrangian multiplier, we get

$$\displaystyle ay + bz = \lambda yz \implies \lambda = \dfrac{ay + bz}{yz}.$$

$$\displaystyle ax + bz = \lambda xz = \dfrac{axy + bxz}{y}.$$

$$\displaystyle bx + by = \lambda xy = \dfrac{axy + bxz}{z}.$$

$$\displaystyle xyz = c \implies z = \dfrac{c}{xy}.$$

Now substitute.

$$\displaystyle ax + bz = \lambda xz = \dfrac{axy + bxz}{y} \implies ax + \dfrac{bc}{xy} = \dfrac{axy + bx * \dfrac{c}{xy}}{y} \implies$$

$$\displaystyle axy + \dfrac{bc}{x} = axy + \dfrac{bc}{y} \implies \dfrac{bc}{x} = \dfrac{bc}{y} \implies x = y.$$

Now for the final substitutions.

$$\displaystyle z = \dfrac{c}{xy} = \dfrac{c}{x^2} \text { and}$$

$$\displaystyle bx + by = \dfrac{axy + bxz}{z} \implies 2bx = \dfrac{ax^2 + bxz}{z} \implies$$

$$\displaystyle 2bz = ax + bz \implies bz = ax \implies \dfrac{bc}{x^2} = ax \implies x = \sqrt[3]{\dfrac{bc}{a}}.$$

Thus, there are indeed three solutions, but only one real solution.

Moreover if a = 600, b = 900, and c = 486

$$\displaystyle x = \sqrt[3]{\dfrac{900 * 486}{600}} = \sqrt[3]{9 * \dfrac{486}{6}} = \sqrt[3]{9 * 81} = 9.$$

$$\displaystyle \therefore y = 9 \text { and } z = \dfrac{486}{9^2} = 6.$$

I am not sure what wolfram did, but I doubt that there are real solutions other than the one found by Mark with far less effort than my nonsense.

EDIT: Unless of course, the method of Lagrangian multipliers is not guaranteed to find all extreme points or I made some idiotic error.

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#### MarkFL

##### Super Moderator
Staff member
I seem to recall running into problems where LM did fail to find critical points, found by AM-GM inequalities instead.

#### JeffM

##### Elite Member
I seem to recall running into problems where LM did fail to find critical points, found by AM-GM inequalities instead.
Interesting. I wonder what the conditions must be for LM to not give all critical points. Time to visit the stack exchange.