Constrained optimization using lagrange multipliers

Question

I was given the following question:

Find the maximum and minimum of the function $f(x,y)=x^2y^2$ subject to $x^2+4y^2=24$.

I tried using lagrange multipliers and got the following:

I was assuming that I would get different answers, and that the biggest one would be the maximum and the smallest would be the minimum. This did not happen. Where did I go wrong, and what steps can I take to solve this problem?

Answer 1

Your function $ \ f(x,y) \ = \ x^2y^2 \ $ has four-fold symmetry about the origin, in that changing the sign of $ \ x \ $ or $ \ y \ $ or both does not alter the value of the function. As the constraint curve, $ \ x^2 \ + \ 4y^2 \ = \ 24 \ $ (an ellipse centered on the origin) also has four-fold symmetry, we should expect to see critical points of a particular type "mirrored" in all four quadrants. Also, since $ \ f(x,y) \ \ge \ 0 \ $ everywhere, we already expect the function to have minima wherever it is equal to zero.

When your Lagrange equations have all of the terms taken to one side of each equation and are factored, we find

$$ 2xy^2 \ - \ 2 \lambda x \ \ = \ \ 2x · (y^2 \ - \ \lambda) \ \ = \ \ 0 \ \ \ , \ \ \ 2x^2y \ - \ 8 \lambda y \ \ = \ \ 2y · (x^2 \ - \ 4\lambda) \ \ = \ \ 0 \ \ . $$

Your calculation for $ \ \lambda \ $ correctly led you to the four absolute maximal points on the ellipse, one in each quadrant, with $ \ f(\pm 2 \sqrt{3} \ , \ \pm \sqrt{3}) \ = \ 36 \ $ .

The other factors in the equations, giving $ \ x \ = \ 0 \ $ or $ \ y \ = \ 0 \ $ then tell us that the absolute minima on the ellipse [ $ \ f(x,y) \ = \ 0 \ $ ] occur at the endpoints of its major and minor axes, $ \ (\pm 2 \sqrt{6} \ , \ 0 ) \ $ and $ \ (0 \ , \ \pm \sqrt{6}) \ . $

$$ $$

(Although it is not asked for here, if we had been asked about critical points in the interior of the ellipse, the first partial derivatives indicate that there is one at the origin. When we work with the second derivatives, however, we receive an inconclusive result (for example, the Hessian determinant is zero there). The problem here is that $ \ f(x,y) \ = \ 0 \ $ everywhere on the coordinate axes, so our usual devices for investigating critical points fail us. We instead need to look at the properties of the function to understand its behavior near the origin.)

Answer 2

Another way to do this is to rewrite the function as $$(24-4y^2)y^2=24y^2-4y^4,$$ which is quadratic in $y^2$ and thus possesses a maximum $$-\frac{24^2-4(-4)(0)}{4(-4)}=36,$$ occurring at the points where $$y^2=-\frac{24}{2(-4)}=3,$$ or where $y=\pm\sqrt 3,$ from where you may find the corresponding values of $x.$