calculuslagrange multiplierpartial derivative
Find the extreme values of the function subject to the given constraint.$$f(x,\, y) = y^2 – x^2,\, x^2 + y^2 = 16$$
I understand how to to compute the extrema using Lagrange multipliers and lambda however I keep getting this question wrong. I end up with
$$-2x=λ*2x$$
$$2y=λ*2y$$
I do not see where you used Lagrange multipliers. Any critical points will satisfy the Lagrange multipliers equation
$$ \begin{bmatrix} yz & xz & xy \end{bmatrix} = \lambda \begin{bmatrix} 2x & 4y & 6z \end{bmatrix} \ . $$
This gives you the system of equations
$$ \begin{align} yz &= 2\lambda x & (1) \\ xz &= 4\lambda y & (2) \\ xy &= 6\lambda z & (3) \\ 96 &= x^2+2y^2+3z^2 & (4) \end{align} $$
First consider $\lambda = 0$ and then consider other cases. Try this and see how it goes for you. I hope this helps.
EDIT:
If you assume $x,y,z\neq 0$, you can solve for $\lambda$ in each of $(1), (2), (3),$ and $(4)$. You then obtain (by equating these $\lambda$), as you did, $6y^2=96$, which gives $y=\pm4$. You similarly obtain $x^2 = 32$, or $x=\pm4\sqrt{2}$, and $z^2 = \frac{32}{3}$, or $z=\pm4\sqrt{\frac{2}{3}}$. With all of the $\pm$'s, you get a few critical points. Plug them in to see which ones are the largest/smallest.
A possible source of confusion here as to the use of the Lagrange multiplier $ \ \lambda \ $ appears to be due to the first Lagrange equation being factored incorrectly. The two equations are found properly, but we should have
$$ 3x^2 \ - \ 2 \lambda \ x \ = \ 0 \ \ \Rightarrow \ \ x \ ( \ 3x \ - \ 2 \lambda \ ) \ = \ 0 \ \ , $$
$$ 8y \ - \ 2 \lambda \ y \ = \ 0 \ \ \Rightarrow \ \ 2y \ ( \ 4 \ - \ \lambda \ ) \ = \ 0 \ \ . $$
So two of the possible solution sets are correctly given as $ \ ( 0 , \ \pm 1 ) \ $ and $ \ ( \pm 1, \ 0) \ $ by applying the constraint equation.
However, there is only the only value for the Lagrange multiplier found from the second equation, $ \ \lambda \ = \ 4 \ $ . This would be applied in the second factor of the first equation to obtain
$$ 3x \ - \ 2 \lambda \ = \ 0 \ \ \Rightarrow \ \ x \ = \ \frac{2 \cdot 4}{3} \ = \ \frac{8}{3} \ \ . $$
This raises an interesting issue in problems of this sort. In this case, the result produced by using the Lagrange multiplier value is inapplicable because there are no points with $ \ x \ = \ \frac{8}{3} \ $ on the constraint surface. [A similar circumstance arises in this problem.] Therefore, the extremal values of our function are
$$ f(0, \ \pm 1) \ = \ 4 \ \ [\text{absolute maximum}] \ \ ; $$
$$ f(1, \ 0) \ = \ 1 \ \ [\text{local maximum}] \ \ ; $$
$$ f(-1, \ 0) \ = \ -1 \ \ [\text{absolute minimum}] \ \ . $$
The "geometric" interpretation of this result is that the values for the function represent maximum and minimum values of the $ \ z-$ coordinate of the possible intersection points of the surface $ \ z \ = \ f(x,y) \ = \ x^3 \ + \ 4y^2 \ $ with the "vertical" cylinder $ \ x^2 \ + \ y^2 \ = \ 1 \ $ .
We show two views of the intersecting surfaces, with three of the four "critical points" visible; $ \ (0, \ -1, \ 4) \ $ is on the "far side" of the diagram at left.
$$ \ \ $$
The value found for the multiplier is the same regardless of the radius of that cylinder, which suggests that the coordinate $ \ x \ = \ \frac{8}{3} \ $ could become important if the constraint surface were "larger". If we were to use the cylinder $ \ x^2 \ + \ y^2 \ = \ 9 \ $ as the constraint "surface", for instance, our result from the Lagrange multiplier would be applicable, now giving us extremal values
$$ f(0, \ \pm 3) \ = \ 36 \ \ [\text{absolute maximum}] \ \ ; $$
$$ f(3, \ 0) \ = \ 27 \ \ [\text{local maximum}] \ \ ; $$
$$ f(-3, \ 0) \ = \ -27 \ \ [\text{absolute minimum}] \ \ ; $$
$$ f\left(\frac{8}{3}, \ \pm \frac{\sqrt{17}}{3}\right) \ = \ \frac{716}{27} \ \approx \ 26.52 \ \ [\text{local minima}] \ \ . $$
For these larger cylindrical radii, the intersection with the surface $ \ f(x,y) \ $ along the "positive $ -x \ $ " face of the cylinder develops a "ripple", leading to the appearance of further "critical points" with $ \ x \ = \ \frac{8}{3} \ $ .
The somewhat more complicated shape of the intersection of the surfaces is visible in the left-hand graph; the critical point $ \ (0, \ -3, \ 36) \ $ is not visible in these views.
Best Answer
Your first equation $-2x=\lambda\cdot 2x$ implies that either $x=0$ or $\lambda=-1$ (or both).
Your second equation $2y=\lambda\cdot 2y$ implies that either $y=0$ or $\lambda=1$ (or both).
Combining those two, you have a total of four possibilities. Look at each possibility and find the possible points (if any) for that possibility. Then look at the function values at each resulting point. Use those to find the maximum and minimum values of your function.