Method of Lagrange multipliers for multiple constraints

Question

I'm having trouble with a past exam question regarding the use of Lagrange multipliers for multiple constraints. The question is:

Using the method of Lagrange multipliers for multiple constraints, find the absolute maximum and minimum of $$f(x,y,z) = xy + 2z$$ On the intersection of $$x + y + z =0$$ And $$x^2 + y^2 + z^2 = 24$$ I'm kinda unsure where to start, I haven't really come across many of these types of questions.

Any help would be appreciated.

Answer 1

You just need to consider $$F=xy+2z+\lambda(x+y+z)+\mu(x^2+y^2+z^2-24)$$ Compute $F'_x,F'_y,F'_z,F'_\lambda,F'_\mu$ and set them equal to $0$.

The same would apply to more constaints. It is just the extension of what you already know and use.

In this particular case where you have one linear constraint, you could eliminate $z$ from it $(z=-x-y)$ and the problem would become $$F=xy-2(x+y)+\lambda(x^2+xy+y^2-12)$$

Answer 2

It seemed like this is a good problem for illustrating the solution of an extremization using variable "elimination" and a single Lagrange multiplier versus the use of two multipliers. Solving the resulting systems proved to be a bit "tricky" in either case.

The function for which we seek extrema, $ \ f(x,y,z) \ = \ xy \ + \ 2z \ , $ has a symmetry under exchanging the coordinates $ \ x \ $ and $ \ y \ $ , so we may expect critical points to be located symmetrically about the plane $ \ y \ = \ x \ $ . The constraint curve is an inclined "great circle" on the surface of a sphere centered on the origin, so this may also produce a limited amount of symmetry.

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In the two-variable, one-multiplier system Claude Leibovici describes, the Lagrange equations are

$$ y \ - \ 2 \ \ = \ \ \lambda · (2x \ + \ y) \ \ \ , \ \ \ x \ - \ 2 \ \ = \ \ \lambda · (2y \ + \ x) \ \ . $$

We will not be able to factor simply the equations we produce by moving all the terms to one side; instead, we consider solving the equations for $ \ \lambda \ : $

$$ \lambda \ \ = \ \ \frac{x \ - \ 2}{2y \ + \ x} \ \ = \ \ \frac{y \ - \ 2}{2x \ + \ y} \ \ . $$

Provided the denominators are not zero (we'll see that they aren't), we "cross-multiply" the ratio to find

$$ 2x^2 \ + \ xy \ - \ 4x \ - \ 2y \ = \ 2y^2 \ + \ xy \ - \ 4y \ - \ 2x \ \ \Rightarrow \ \ x^2 - x \ = \ y^2 - y \ \ . $$

(The challenge in solving systems of non-linear equations is in finding an evident path to a solution -- when there is such a path; I'll admit that it wasn't clear to me right away how to continue. In fact, I didn't see it until I'd already solved the two-multiplier system.)

$$ \Rightarrow \ \ y^2 - x^2 \ = \ y - x \ \ \Rightarrow \ \ (y + x) · (y - x) \ \ = \ \ y - x \ \ \Rightarrow \ \ (y + x - 1)·(y - x) \ \ = \ \ 0 \ \ . $$

[PROTIP: Resist the urge to simply "cancel" like factors on either side of an equation; you may neglect possible solutions to the equation or create false ones in doing so.]

So we have the two cases, $ \ y \ = \ x \ $ (as anticipated) and $ \ y \ = \ 1 - x \ . $ With the constraint equation written as $ \ x^2 \ + \ xy \ + \ y^2 \ = \ 12 \ , $ we find

$$ \ \mathbf{y = x : } \quad x^2 \ + \ x·x \ + \ x^2 \ = \ 12 \ \ \Rightarrow \ \ 3x^2 \ = \ 12 \ \ \Rightarrow \ \ x \ = \ y \ = \ \pm 2 \ \ , \ \ z \ = \ -(x+y) \ = \ \mp 4 \ \ ; $$

$$ \ \mathbf{y = 1-x : } \quad x^2 \ + \ x·(1-x) \ + \ (1-x)^2 \ = \ x^2 \ - \ x \ + \ 1 \ = \ 12 \ \ \Rightarrow \ \ x^2 \ - \ x \ - \ 11 \ = \ 0 $$ $$ \Rightarrow \ \ x \ = \ \frac{ 1 \ \pm \ \sqrt{ \ 1 \ - \ 4·1·(-11)}}{2} \ = \ \frac{1 \ \pm \ \sqrt{45}}{2} \ \ \text{or} \ \ \frac{1 \ \pm \ 3\sqrt{5}}{2} \ \ , \ \ y \ = \ \frac{1 \ \mp \ 3\sqrt{5}}{2} \ \ , $$ $$ z \ = \ - \left( \frac{1 \ \pm \ 3\sqrt{5}}{2} \ + \ \frac{1 \ \mp \ 3\sqrt{5}}{2} \right) \ = \ -1 \ \ . $$

The two extremal values from the $ \ y = x \ $ case yield $ \ f( \ 2 \ ,\ 2 \ , \ -4 \ ) \ \ = \ \ 2·2 \ + \ 2·(- 4) \ = -4 \ $ and

$$ \ f( \ -2 \ ,\ -2 \ , \ +4 \ ) \ \ = \ \ (-2)·(-2) \ + \ 2·4 \ \ = \ \ 12 \ \ , $$

the maximum value for our function. From the second case, we obtain the minimum value,

$$ \ f \left( \frac{1 \ \pm \ 3\sqrt{5}}{2} \ , \ \frac{1 \ \mp \ 3\sqrt{5}}{2} \ , \ -1 \right) \ \ = \ \ \left( \frac{1 \ \pm \ 3\sqrt{5}}{2} \right)·\left( \frac{1 \ \pm \ 3\sqrt{5}}{2} \right) \ + \ 2·(-1) $$

$$ = \ \ \frac{1}{4} \ - \ \frac{45}{4} \ - \ 2 \ \ = \ \ \frac{-52}{4} \ \ = \ \ -13 \ \ . $$

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We can also work with all three coordinate variables and two multipliers to produce the system of equations

$$ y \ = \ \lambda · 1 \ + \ \mu · 2x \ \ , \ \ x \ = \ \lambda · 1 \ + \ \mu · 2y \ \ , \ \ 2 \ = \ \lambda · 1 \ + \ \mu · 2z \ \ . $$

Solving each of these for $ \ \lambda \ $ gives us

$$ \lambda \ \ = \ \ y \ - \ 2 \mu x \ \ = \ \ x \ - \ 2 \mu y \ \ = \ \ 2 · (1 \ - \ \mu z) \ \ . $$

The first pair-equation can be re-written as $ \ y \ + \ 2 \mu y \ - \ x \ - \ 2 \mu x \ = \ 0 \ \ \Rightarrow \ \ (y - x) · ( 1 \ + \ 2 \mu) \ = \ 0 \ \ , $ leading us to the two cases $ \ y \ = \ x \ $ and $ \ \mu \ = \ -\frac{1}{2} \ . $

For $ \ y \ = \ x \ , $ we obtain $ \ z \ = \ -(x+y) \ = \ -2x \ \ \Rightarrow \ \ z^2 \ = \ 4x^2 \ = \ 4y^2 \ . $ Putting this into the spherical constraint yields $$ \ x^2 \ + \ x^2 \ + \ 4x^2 \ = \ 24 \ \ \Rightarrow \ \ x^2 \ = \ y^2 \ = \ 4 \ \ , \ \ z^2 \ = \ 16 $$

and the relations here imply $ \ x \ = \ y \ = \ \pm 2 \ \ \ , \ \ z \ = \ \mp 4 \ , \ $ as above.

Using $ \ \mu \ = \ -\frac{1}{2} \ $ in the $ \lambda-$ equation produces $ \ \lambda \ = \ x \ + \ y \ = \ 2 \ + \ z \ \ . $ Inserting this into the planar constraint then gives us $ \ x \ + \ y \ + \ z \ = \ ( 2 + z) \ + \ z \ = \ 0 \ \Rightarrow \ z \ = \ -1 \ , \ $ and consequently, $ \ x \ + \ y \ = \ 1 \ \Rightarrow \ y \ = \ 1 \ - \ x \ . \ $ The spherical constraint equation becomes

$$ x^2 \ + \ (1 \ - \ x)^2 \ + \ (-1)^2 \ \ = \ \ 24 \ \ \Rightarrow \ \ x^2 \ + \ 1 \ - \ 2x \ + \ x^2 \ + \ 1 \ \ = \ \ 24 $$ $$ \Rightarrow \ \ 2x^2 \ - \ 2x \ - \ 22 \ \ = \ \ 0 \ \ , $$

which is essentially the quadratic equation we found earlier.