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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?

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    $\begingroup$ Yes, but we can't tell you what, since you don't show us your computations. $\endgroup$ – Jan Aug 26 '19 at 14:43
  • $\begingroup$ @Jan edited the post. $\endgroup$ – user562600 Aug 26 '19 at 14:50
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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.

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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!

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