# Lagrange multipliers to find maximum and minimum value

Use Lagrange multipliers to find the maximum and minimum values of f(x; y) = x^2+4y^3 subject to the constraint x^2 + 2y^2 = 8. Also, find the points at which these extreme values occur.

Using Lagrange multipliers, we get, 2x = λ2x

12y^2 = λ4y

From the first equation, we get λ=1, putting in the second equation we get y=1/3, 0. Using these two values of y and constraint equation, we get x = √70/3,2√2) respectively. Thus the points at which maxima and minima occur are (√70/3,1/3) and (2√2,0) but the actual minima is at (0,-2). Am I doing something wrong?

• Yes, but we can't tell you what, since you don't show us your computations. – Jan Aug 26 at 14:43
• @Jan edited the post. – redx Aug 26 at 14:50

The Lagrange-function of the system is

$$\mathcal{L}(x, y, \lambda) = x^2 + 4y^3 - \lambda (x^2 + 2y^2 - 8).$$

In order to get the stationary points one has to use the necessary condition $$\nabla{\mathcal{L}} = 0$$, which leads to the equation system

\begin{align} 2x - 2\lambda x &= 0 \\ 12 y^2 - 4\lambda y &= 0 \\ x^2 + 2y^2 &= 8. \end{align}

The first equation is equivalent to $$x ( 1- \lambda) = 0$$, so $$x = 0$$ or $$\lambda = 1$$. Plugging $$x = 0$$ into the last equation gives us $$2y^2 = 8$$, from which we conclude the stationary points $$(0, \pm 2)$$.

With $$\lambda = 1$$ we get in the second equation $$12y^2 - 4y = y( 12 y - 4) = 0$$, i. e., $$y= 0$$ or $$y = \frac{1}{3}$$. Plug this into the third equation, then you get more stationary points. Now you have to prove a sufficient condition in order to get their type.

from 2x = λ2x you also get x=0 and λ anything, then the constraint gives y^2=4, and f_y=0 -> λ≠1 all three equations must be used!