# 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

Solve for y^2: y^2 = 16 - x^2 ===> f(x,y) = 2x^2 + 3(16 - x^2) - 4x - 5 = -x^2 - 4x + 43 = -(x + 2)^2 + 47 = g(x).

Consider f(x,y) = g(x) = 47 - (x + 2)^2 on [-4, 4]. Clearly: g(x) <= 47 and that max g(x) = 47 when x = -2, so y^2 = 16 - (-2)^2 = 12 , and y = 12^(1/2) and y = - 12^(1/2). Observe that g is minimized when (x + 2)^2 is maximized, and this occurs when x = 4. So min g = 47 - (4 + 2)^ = 47 - 36 = 11.

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Thank you for the help! Makes sense now. –  Elephant Mar 7 '14 at 1:58