# Use Lagrange multipliers to find the maximum(s) and minimum(s)

Use Lagrange multipliers to find the maximum(s) and minimum(s) of

$$f(x,y)=2x^2+3y^2−4x−5 \text{ subject to } x^2+y^2=16$$

So far I've taken the partials of f and g:

\begin{align} f_x &= 4x-4 \\ f_y &= 6y \\ g_x &= 2x \\ g_y &= 2y \end{align}

Then I set them equal to each other and multiplied by $\lambda$: \begin{align} 4x-4 &= 2x\lambda \\ 6y &= 2y\lambda \end{align}

I'm kind of stuck here where I need to solve for $\lambda$.

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just use partial derivatives – Knight Mar 7 '14 at 0:50
If $6y = 2y\lambda$, what's $\lambda$? – dfan Mar 7 '14 at 1:08
3, so would that mean x = -2? Also, if that is correct, do you know how I could solve for y? thanks – Elephant Mar 7 '14 at 1:10
Don't forget that you have three constraints to find your three variables $x$, $y$, and $\lambda$: the two equations you got from equating partial derivatives, and the original constraint $x^2+y^2=16$. – dfan Mar 7 '14 at 1:12
In Lagrange multiplier problems it is vital to find all solutions of your equations. So $6y=2y\lambda$ does not tell you that $\lambda=3$, it tells you that either $\lambda=3$ or $y=0$. – David Mar 7 '14 at 1:13