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Answer 1

It's on Lagrange multipliers.

Answer 2

Are you sure you recorded the problem correctly? The first condition gives \[ \lambda =1\], but the second gives \[ \lambda = 2 \]

Answer 3

very. It's a system of equations copied directly from the textbook in front of me

Answer 4

Well, those gradients will never be equal which is what Lagrange multipliers is searching for... when two gradients are equal. One gradient is {2 x, 2 y} and the other is {2 x, 4 y}. Trying to find a point where they will be equal is impossible. hmm...

Answer 5

However, you don't need lagrange multipliers... you can just substitute and do the normal derivative maximization to find the x.

Answer 6

there is also the condition of the circle which in this case would act like the Closed Interval Method where the circle is the endpoints

Answer 7

true, but this section is on Lagrange

Answer 8

Ok, let's try to figure this out. What do you mean by the "Closed Interval Method" By this do you mean find the possible x's that satisfy the circle (\[ -1 \le x \le 1\]) then find the maximum of the function on this interval?

Answer 9

well, the answer in my book is that the max occurs when f(0,+/-1)=2 and the min occurs when f(+/-1,0)=1 so in this case lagrange is really overkill, but essentially, since this is a function of two variables and an equation of three variables, it's not surprising that it is in 3 dimensions