# Extreme values (Lagrange multipliers)

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 $$\nabla$$f=$$\lambda$$$$\nabla$$g ,which can be written as

$$f_{x}$$=$$\lambda$$$$g_{x}$$

$$f_{y}$$=$$\lambda$$$$g_{y}$$

g(x,y)=36

or as

y=$$\lambda$$8x

x=$$\lambda$$18y

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

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

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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.

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

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.

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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:
$$(x,y) = (\sqrt{\frac{18}{5}}, - \frac{2}{3}\sqrt{\frac{18}{5}})$$
$$(x,y) = (- \sqrt{\frac{18}{5}}, \frac{2}{3}\sqrt{\frac{18}{5}})$$