Extreme values (Lagrange multipliers)

[aicort]

determine, if any, the maximum and minimum values of the scalar field f (x, y) = xy subject to the constraint 4x^2{}+9y^2{}=36


The attempt at a solution

using Lagrange multipliers, we solve the equations \nablaf=\lambda\nablag ,which can be written as

f_{x}=\lambdag_{x}

f_{y}=\lambdag_{y}

g(x,y)=36

or as

y=\lambda8x

x=\lambda18y

4x^2{}+9y^2{}=36

it's pretty much all done but can somebody solve this? cause i have some doubts about which are the extreme points

 

[Coto]

I would say your g(x,y) = 36 is a little suspicious. If this were the case then the two statements above would be false since g_x and g_y are 0.

 

[aicort]

no, in fact g(x,y)=4x^2{}+9y^2{}=36 but well, I wrote it that way because is the way my book does

 

[Fightfish]

We already know that (0,0) is a turning point which has a value f (0,0) = 0.

Leaving that aside for now, solving the set of equations that you listed:
y = 8\lambda x
x = 18\lambda y
4x^2 + 9y^2 = 36
would yield one or more points.

Then find the value of f(x,y) at each of these points and compare.

 

[aicort]

is it correct?

http://img245.imageshack.us/img245/4636/lagrange.th.jpg

 

[Fightfish]

Firstly, careless mistake: y is +/- \frac{2}{3}\sqrt{\frac{18}{5}} as you've written on the third line, but in carrying it over into f(x,y), you mixed up the denominator and numerator.

Secondly, \lambda = +/- \frac{1}{12}, so there exist two other solutions:
(x,y) = (\sqrt{\frac{18}{5}}, - \frac{2}{3}\sqrt{\frac{18}{5}})
(x,y) = (- \sqrt{\frac{18}{5}}, \frac{2}{3}\sqrt{\frac{18}{5}})