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)

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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)

yesNormally, the constraint is an equation...should it be:

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

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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.

At least to me, this was far from obvious.\(\displaystyle 600y+900z=\lambda(yz)\)

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

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

The first two equations imply \(x=y\)

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

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

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

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

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

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.

Last edited:

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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.

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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.

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

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

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

https://mathoverflow.net/questions/334138/limitations-on-method-of-lagrangian-multipliers

LaGrange multipliers will get you all critical points for sufficiently smooth functions. Of course, you still must test for extrema at the boundaries.

Because the entire problem is perfectly symmetric with respect to x and y, the additional solution provided by wolfram is spurious.

This gets us right back to Mark's first post.