Finding minimum and maximum points with Lagrange Multipliers

Question

I need to use Lagrange Multipliers to find the maximum and minimum values of the function:

$f(x,y)=2e^{xy}$

subject to the given constraints:

$2x^2+y^2=32$

So I went through some examples, and I got:

$x=\pm2\sqrt{2}$ and $y=\pm4$ (Wolfram confirms).

Now I'm having trouble finding the maximum and the minimum. I understood that If I want to find out if $(2\sqrt2,4)$ is maxima or minima, then I'll take for example $x=3$, and by the constraint, this point will be $(3,\sqrt{14})$, and check if this value is greater or smaller than (for example) $(2\sqrt2,4)$, and if it is bigger - then my $(2\sqrt2,4)$ is a minimum.

But in my example, $3>2\sqrt2$ and $4>\sqrt{14}$, isn't it problematic?

This seems a bit messy. First, am I right? Second, any other way to do so, that is not much complicated?

Answer 1

The points found are guaranteed to be maxima or minima, so it's enough to plug each pair back into $f(x)$ to determine which are which

Doing the way you suggested is not problematic, it's only required that the points are on the contraint, an ellipse in this case. If you plug, say, $(2\sqrt{2},4)$ and $(3,\sqrt{14})$

$$2e^{8\sqrt{2}}\approx 163874.4$$

$$2e^{3\sqrt{14}}\approx 149959.3$$

So it doesn't matter that $3>2\sqrt{2}$ because $\sqrt{14} < 4$. The process of Lagrange multipliers sorted it out for you already by finding the values that beat all the others

Answer 2

I didn't check the calculation, but in general, once we have obtained the possible max/min points by Lagrange Multipliers, we simply need to plug the values into $f(x,y)$ and determine the max/min values, that is:

• $f(2\sqrt 2,4)=2e^{8\sqrt 2}=\dots$
• $\dots$

In this case it is not difficult to see that

• $f_{max}=f(2\sqrt 2,4)=f(-2\sqrt 2,-4)=2e^{8\sqrt 2}$

• $f_{min}=f(-2\sqrt 2,4)=f(2\sqrt 2,-4)=2e^{-8\sqrt 2}$

Answer 3

You don't really need to "plug" any numbers into the function in order to answer your question. The function $ \ f(x,y) \ = \ 2e^{xy} \ $ has symmetry about the origin [ $ \ f(x,y) \ = \ f(-x,-y) \ $ ] . The constraint curve $ \ 2x^2 \ + \ y^2 \ = \ 32 \ $ is an ellipse centered on the origin, so we expect a critical point in any quadrant to be "mirrored" by a critical point of the same kind in the diagonally-opposite quadrant.

You found two critical values for $ \ x \ $ and $ \ y \ $ each, producing four critical points in all, one in each quadrant as anticipated. The product $ \ xy \ $ is positive in the first and third quadrants and negative in the second and fourth. Thus, the exponent in $ \ 2e^{xy} \ $ is positive at $ \ (2\sqrt2 \ , \ 4) \ $ and $ \ (-2\sqrt2 \ , \ -4) \ $ , producing the absolute maxima for our function, with the exponent being negative at $ \ (2\sqrt2 \ , \ -4) \ $ and $ \ (-2\sqrt2 \ , \ 4) \ $ , where the absolute minima are located. (The values are as shown by user.)