# Lagrange multipliers

1. Aug 1, 2010

### ForMyThunder

1. The problem statement, all variables and given/known data
Using Lagrange multipliers, find the maximum and minimum values of $$f(x,y)=x^3y$$ with the constraint $$3x^4+y^4=1$$.

2. Relevant equations

3. The attempt at a solution
Here is my complete solution. I just wanted to make sure there are no errors and I did it correctly. Thanks for any feedback.

$$\nabla f = \lambda \nabla g$$
$$3x^2y=12\lambda x^3$$ and $$x^3=4\lambda y^3$$
Solving these I got $$\lambda = \frac{1}{4}$$ and $$\lambda = -\frac{1}{4}$$

Putting these values into the equation on the right gives x=y and x=-y. Substituting these into the left equation gives $$x=y=\frac{1}{\sqrt{2}}$$ and $$x=\frac{1}{\sqrt{2}}$$, $$y = -\frac{1}{\sqrt{2}}$$.

Putting these values into the equation for $$f$$ gives a maximum of $$\frac{1}{4}$$ and minimum of $$-\frac{1}{4}$$.

Last edited: Aug 1, 2010
2. Aug 1, 2010

### arildno

Your maximum and minimum values are correct, but you have, for the maxima two soluitons,
$$x=y=\pm\frac{1}{\sqrt{2}}$$
rather then just one maximum.

Similarly for the minimum value.