you just apply the method for once got the max. or min. points, keep those which satisfy
This normally wouldn't be so hard for me, except for the awkward constraint we've been given.
Find the maximum and minimum values of the function
when is a point in the set
I've been told that using Lagrange Multipliers is the proper method to solve this, but I'm not so sure. I'm new to using Lagrange Multipliers, and I haven't seen a question for which the constraint is a set that doesn't have a definite value.
I'm not sure what Krizalid is saying. If you "apply the method" (Lagrange multipliers) for , any points you get will satisfy and so .
Here's what I would do. Find all x and y such that , that is, such that and . Note those (x, y) that satisfy .
Then use Lagrange multipliers to find those (x, y) that maximize or minimize f with the constraint .
Finally, check the value of f at each of those (x, y) points, both inside and on , to determine the maximum and minimum values of f.
HallsofIvy, either I did your method wrong, or something else happened that I'm not familiar with. I've not seen the method you're using in our course material yet.
When I went for and , I end up with and , which leads to in both cases.
Did I do this incorrectly somehow, or this some sort of a "trick question" thing?
EDIT: Okay, using the Lagrange method of seems to either be malfunctioning on me, or I'm doing it wrong. Again, our source textbook is next-to-worthless.
Using your constraint , I have . Taking partial derivatives gets me and . Finding , however, gets me , which I'm guessing is a mistake of some sort.
Is this method incorrect?
so . At any max or min in the interior of the set we must have so one of x= 0, x=y, x=-y must be true. Also so one of y= 0, , or must be true.
If x= 0, only y= 0 satisfies the other equations. If y= x again we have only x= y= 0. I don't see how x= -1/2, y= -1/5 satisfy those equations. I find only x= y= 0.
On the boundary, , so we must have or and .
It's never necessary to find since that is not relevant to the solution. Instead, eliminate by dividing one equation by the other:
That, together with gives two equations to solve for x and y.
According to this result I got off WolframAlpha, however, I have the following max/min values:
See this link
WolframAlpha isn't 100% reliable at times, but I calculated using the max roots and it worked out to equal 80. However, I'm wondering if the program misinterpreted what I put in.
Can someone confirm which of my answers would be correct?