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I'm having trouble figuring out the critical points using the Lagrange Multipliers method for the following exercise:

Using the method of Lagrange multipliers, determine the critical points of the function $f(x,y) = x^2+24xy+8y^2$ subject to the constraint $g(x,y) = x^2+y^2−25 = 0$

My work was as follows:

$$F = (x^2+24xy+8y^2) - \lambda(x^2+y^2-25)$$ $$F_x = 2x + 24y -2\lambda x = 0$$ $$F_y = 24x+16y-2\lambda y =0$$ $$F_{\lambda} = -x^2 - y^2 + 25 =0$$

Now, from here, I ran into problems trying to find $x,y$ and $\lambda$. I tried first substituting and solving:

From equation 1: $$x = \frac{-24y}{(2-2\lambda)}$$ $$\implies y = 24\left(\frac{-24y}{(2-2\lambda)}\right)+16y-2\lambda y=0$$

And ended up with:

$$\left(\frac{-576y}{(2-2\lambda)}\right)+16y-2\lambda y =0$$ $$y\left(\left(\frac{-576}{(2-2\lambda)}\right) + 16 - 2\lambda)\right)=0$$

And.. here's where I get stuck, if this is even the right route. I can get an answer for lambda right now and have $y=0$ as a solution, making $x=0$ one as well, although I'm pretty sure that's wrong as is. Also, I haven't even used the third equation, which is usually where I find $\lambda$ and then subsequently $x,y$. What's going on here? How should I solve this?

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  • $\begingroup$ Your final equation is fine, just solve the quadratic in $ \lambda$. $\endgroup$ – Donald Splutterwit Dec 12 '17 at 23:55
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I think you can get an answer for $\lambda$ from the equation you have, which gives $\lambda_1 = -8$, $\lambda_2 = 17$. Then if you put it in the equation that you found $x$, only by using the equation $x^2+y^2=25$ you can solve that one.

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Your final equation is fine, just solve the quadratic in $ \lambda$. A slightly easier way to this equation is to rewrite your $F_x$ and $F_y$ equations as \begin{eqnarray*} 12 y = x(\lambda-1) \\ 12x=y(\lambda-8) \end{eqnarray*} Now multiply these equations together (and assume $xy \neq 0$) and we have \begin{eqnarray*} (\lambda-1) (\lambda-8)=12^2. \end{eqnarray*} You can complete the solution from here ?

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  • $\begingroup$ Uh, think so? Once I have the values for lambda, I can perhaps find $y$ and $x$ using an augmented matrix found from the two equations you gave me? $\endgroup$ – sangstar Dec 13 '17 at 0:09
  • $\begingroup$ Take one of the solutions eg $\lambda=17$, both the equations give $3y=4x$, now sub this into $x^2+y^2=25$, etc ... $\endgroup$ – Donald Splutterwit Dec 13 '17 at 0:13
  • $\begingroup$ Ah, so to solve for $x$, for example, given $\lambda = 17$, I'd do $x^2 + (4x/3)^2 = 25$? $\endgroup$ – sangstar Dec 13 '17 at 0:45
  • $\begingroup$ That's right ... giving $x= \pm 3 , y= \pm 4$ ... now do the same again for the other value of $\lambda$. $\endgroup$ – Donald Splutterwit Dec 13 '17 at 0:49

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