0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.