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I want to solve a Lagrange multiplier problem,

$$f(x,y) = x^2+y^2+2x+1$$ $$g(x,y)=x^2+y^2-16 $$

Where function $g$ is my constraint. $$f_x=2x+2, \ \ \ f_y=2y, \ \ \ g_x=2x\lambda, \ \ \ g_y=2y\lambda$$

$$ \begin{cases} 2x+2=2x\lambda \\ 2y=2y\lambda \\ x^2+y^2-16=0 \end{cases} $$

See, this is a very nasty system of equations. At any rate, I get $\lambda = 1$ because in this case, $y=0$. So I cannot do anything with this as far as algebra is concerned? How do I resolve a problem like this?

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    $\begingroup$ The last line $x^2+y^2-16$ is not an equation. $\endgroup$
    – edm
    Sep 30, 2017 at 2:30
  • $\begingroup$ @edm: It is, now. You could have been more welcoming and solve the issue yourself. $\endgroup$ Sep 30, 2017 at 2:43
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    $\begingroup$ Why use Lagrange multipliers? If $g(x,y)=0$ then $f(x,y)=2x+17$, so one is just maximising/minimising $x$ on the circle with centre $(0,0)$ and radius $4$. Obviously the extreme points are $(x,y)=(\pm4,0)$. $\endgroup$ Sep 30, 2017 at 3:41
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    $\begingroup$ Lagrange multiplier problems are notorious for highlighting sloppy algebra. The desired answers are often lost when you do common mistakes like divide by a variable that could be zero, which I imagine is how you arrived at $\lambda = 1$. $\endgroup$
    – user14972
    Sep 30, 2017 at 7:25

2 Answers 2

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$2x(1-\lambda) = -2\tag 1$

$2y(1-\lambda) = 0\tag 2$

From (2) Either $y = 0$ or $(1-\lambda) = 0$

$(1-\lambda) \ne 0$ because if it were (1) would not be true

Thus $y = 0$

Plug in the value of y in g(x,y) and find x.

and $x = +/- 4$

The points are $(4,0)$ and $(-4,0)$

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  • $\begingroup$ Can you work this entire solution out? How are you getting $\pm4$? $\endgroup$
    – Computer
    Sep 30, 2017 at 5:08
  • $\begingroup$ plug the value of y=0 in g(x,y) and you will get $x^2 = 16$ and hence the solution $\endgroup$ Sep 30, 2017 at 5:11
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this method was taught in our class Hope this could help you

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  • $\begingroup$ What do you mean nothing can be predicted? That is not an option to not be able to predict the local maximum or minimum $\endgroup$
    – Computer
    Sep 30, 2017 at 7:40
  • $\begingroup$ Since Hessian matrix is zero. We can't predict whether $f(x,y)=x^2+y^2+2x+1$ will maximize or minimise under given constraint $x^2+y^2-16$ $\endgroup$
    – ashi
    Sep 30, 2017 at 7:59

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