Lagrange Multiplier when one variable is equal to zero?

Question

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?

Answer 1

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

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

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 $(-4,0)$

Answer 2

this method was taught in our class Hope this could help you