Re: maximization problem
- From: "Stephen J. Herschkorn" <sjherschko@xxxxxxxxxxxx>
- Date: Wed, 06 Jun 2007 00:29:32 -0400
chrizm7@xxxxxxxxx wrote:
On Jun 5, 6:58 pm, "Stephen J. Herschkorn" <sjhersc...@xxxxxxxxxxxx>
wrote:
chri...@xxxxxxxxx wrote:
Find the maximum of f(x1, ..., x_n) = (x1*x2*...*x_n)^2 subject to the constraintLetting y_i = (x_i)^2, your problem is
g(x1,...,x_n) = x1^2 + ... + x_n^2 = 1
I try to use lagrange multipliers. I take grad(f) and grad(g) and set
grad(f) = \lambda*grad(g)
but I always get stuck solving the nonlinear system of equations. Is there a trick to doing this?
max sum(i=1..n, ln y_i)
s.t.
sum(i=1..n, y_i) = 1
y_i > 0 for all i
I suspect one could come up with some non-Lagrange argument that y_i = 1/n for all i should be the answer.
Okay, so f(y1, ..., y_n) = ln(y1*...*y_n) = ln(y1) + ... + ln(y_n).
g(y1,...,y_n) = y1 + ... +y_n = 1
Taking the gradients I get the vector equation:
(1/y1, ..., 1/y_n) = \lambda*(y1, ..., y_n)
Umm, what's that? Check your right-hand side.
I dot product both sides with (y1, ..., y_n) to get:
Now why would you do that? Your last vector equation gives you n scalar equations; why collapse them? Even with the erroneous equation, you should end up with the same answer.
n = \lambda*||y||^2
\lambda = n / ||y||^2
Plugging lambda back in I get
(1/y1, ..., 1/y_n) = n / ||y||^2*(y1, ..., y_n)
So 1 / y_i = n / ||y||^2*y_i
And even that implies y_i = y_j for all i, j.
How did you get y_i = 1/n for all i from that?
--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan
.
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