kostoglotov said:
Homework Statement
My question is quite specific, but I will include the entire question as laid out in the text
Consider the problem of minimizing the function f(x,y) = x on the curve y^2 + x^4 -x^3 = 0 (a piriform).
(a) Try using Lagrange Multipliers to solve the problem
(b) Show that the minimum value is f(0,0) = 0 but the Lagrange condition \nabla f(0,0) = \lambda \nabla g(0,0) is not satisfied for any value of \lambda
(c) Explain why Lagrange Multipliers fail to find the minimum values in this case
Homework Equations
The Attempt at a Solution
[/B]
I've answered (c) correctly, but I'm not happy with my own answer, because I don't really understand why it's correct.
I arrived at the answer by plotting f(x,y) = x in Matlab with contour curves of the constraint and then zoomed in on the contour curve where it equals 0.
I got this:
It's hard to see, but where that red ring is, is (0,0,0), which is the constrained min of f(x,y). So I can see graphically that my constraint is discontinuous at (0,0).
The solutions manual to the text gives the answer as "\nabla g(0,0) = 0 and one of the conditions of the Lagrange method is that \nabla g(x,y) \neq 0".
Ok, so a condition of the method is that the grad vector of the constraint not be a zero vector. But why?
I tried solving the general form of the constraint as a limit as (x,y) approach (0,0) but couldn't get an answer. Yet I can clearly see on the graph that the constraint as a level curve at g(x,y) = 0 is discontinuous at (0,0).
If I hadn't had Matlab available, I wouldn't have been able to answer this question. How could I have approached it analytically?
In order for the Lagrange multiplier method be a provably valid necessary (not sufficient) condition for a max or min of ##f(x,y)## at ##(x_0,y_0)##, it is necessary to satisfy some type of constraint qualification at that point. There are several types, but the simplest and most "computable" qualification is that the gradient vectors of the constraint functions be linearly independent at there. For a single constraint ##g(x_1, x_2, \ldots, x_n) = 0## that means that ##\nabla g \neq \vec{0}## at the point ##p_0 = (x_{10}, x_{20}, \ldots , x_{n0})##.
If you do not have a constraint qualification, the Lagrange multiplier rule
may, or may not hold; it just cannot be guaranteed by a theorem.
Basically, if you have a nonzero gradient, you can think of solving for one of the variables in terms of the others in a neighborhood of your "optimum" ##p_0##. So, eg,, if ##g_1 \neq 0## (where ##g_i = \partial g/ \partial x_i |_{p_0}##) then along the constraint ##g = 0## we have
dx_1 = -\frac{g_2}{g_1}\, dx_2 - \ldots - \frac{g_n}{g_1}\, dx_n
so (with ##f_i = \partial f/ \partial x_i |_{p_0}##) we have
df(\vec{x})|_{\vec{x} = p_0} = f_1\, dx_1 + f_2 \, dx_2 + \cdots + f_n \, dx_n \\<br />
= \left( f_2 - \frac{f_1}{g_1} g_2 \right) \,dx_2 + \cdots + \left(f_n - \frac{f_1}{g_1} g_n \right)\, dx_n \\<br />
= (f_2 - \lambda g_2)\, dx_2 + \cdots + (f_n - \lambda g_n)\, dx_n
where ##\lambda = f_1 / g_1##. Now, in a neighborhood of ##p_0## we can regard the variables ##x_2, \ldots, x_n## as independent, so we need the above ##df = 0##. Those give components ##2, \ldots, n## of the Lagrange conditions. Component 1 holds as well, since we can re-write the definition of ##\lambda## in the form ##f_1 - \lambda g_1 = 0##.
Note that there is more involved here than just the "geometry" of the constraint set; it matters how what geometrical object is represented. For example, take the problem
\min x^2 + y^2, \;\; \text{subject to } \; \; x + y = 1.
The obvious solution is at (1/2,1/2), and the Lagrange multiplier rule holds.
Now suppose that instead of representing the line as ##x+y-1 = 0## we (stupidly) represent it as ##g_1(x,y) = (x+y-1)^2 = 0##. We get exactly the same ##(x,y)## point set in both cases, but now the problem
\min x^2 + y^2, \;\ \text{subject to} \;\ (x+y-1)^2 = 0
fails to satisfy the Lagrange multiplier rule. At the optimum (1/2,1/2) there is NO Lagrange multiplier ##\lambda##.