using the method of LaGrange multipliers find the extreme values of the function
$f(x,y)=x^3+4y^2$ with constraint $x^2+y^2=1$
currently I have that: $3x^2=2Lx$ which leads to $x=0$ or $L=3/2$
and: $8y=2Ly$ which leads to $L=4$ or $y=0$
using the constraint when $x=0$ $y=+/-1$ which leads to $f(0,+/-1)=4$
using the constraint when $y=0$ $x=+/-1$ which leads to $f(+/-1,0)=1$
where do I go from here it seems that $f(0,+/-1)=4$ is a maximum and $f(+/-1,0)=1$ is a minimum, and what is the significance of L since we never use it? any help would be greatly appreciated.
Best Answer
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.