Question

Maximize the function f(x,y) = x- y^2 subject to the constraint 2x + y^2 = 4 . Question says use the Lagrange method.
I tried to use it :

fx = λgx
fy = λgy

2x + y^2 = 4
After derivatives I got:

1 = λ2
-2y = λ2y

λ = 1/2 and λ = -1 ??? Is there a problem with that question or am I making a simple mistake because I was solving questions for 5-6 hours without taking a break and that question made me crazy. In examples in my notes, we found only one λ and solution was easy. I couldn't find any sample example like that so please help me.

Answer 1

I know how to solve this but I am stuck because of λ values :(

Answer 2

λ can not be -1 because it fails to satisfies the equation 1 = 2λ

Answer 3

Ok what it's saying is -2y (key letter is the y which is an exponent) so essentially it's saying that said value times 2y is equivalent to -2y so y must equal -1/2

Answer 4

or positive \[\frac{ 1 }{ 2}\] or something to that effect sorry

Answer 5

But as @niko said it can't be -1 because it doesn't satisfy the equation but indirectly it's telling you that (I can't find the sign on my computer) is equal to \[\frac{ 1 }{ 2 }\]

Answer 6

but 1/2 satisfy first equation and it fails at second equation so how can I choose which one to use?

Answer 7

no, 1/2 doesn't fail to satisfy the equation -2y = 2yλ.

What you did wrong was you assume y to be non-zero. That's why you get λ = -1

Answer 8

there is no restriction on the values of y. In fact, it's implied that y is 0, not assumed to be 0

Answer 9

So, If we consider y = 0, λ = 1/2 can satisfy both equations. Now, will I continue solving this problem with accepting λ = 1/2 ?

Answer 10

not quite sure what you meant, but λ = 1/2 from the first equation. This implies y = 0. And now that we know y = 0, we get x = 2.

so the only solution you get is
(x,y) = (2,0)

Answer 11

Oh, so since we have only 1 solution, can we say that point maximize the equation?

Answer 12

how can we be sure that it is maximum and not minimum?

Answer 13

That's a good question. So we need another point to compare but how can I obtain another point value?

Answer 14

you use partial derivative test.

Answer 15

I have no idea how to use it right now

Answer 16

http://tutorial.math.lamar.edu/Classes/CalcIII/RelativeExtrema.aspx

scroll down where you see "Fact" above the green box

Answer 17

Oh I remember now. I need to check D<0 or D>0 to find out max or min point.Am I right?

Answer 18

yes

Answer 19

Okay I will work on it in the morning because it's almost 3 am here. Thank you so much for your help!

Answer 20

no pro