Use Lagrange multipliers to find the maximum and minimum values of the function :$$f(x,y)=e^{xy}$$ constraint $$x^3+y^3=16$$

This is my problem in my workbook. When I solve, I'm just have one solution, so I cannot find other. Here is my solution:

$$f(x,y) = e^{xy}$$ $$g(x,y) = x^3+y^3-16$$ $t(x,y) = f(x,y) + \lambda*g(x,y)$ So we will have three equations by Lagrange Multiplier:

$$(1) y*e^{xy} + 3*\lambda*x^2 = 0$$ $$(2) x*e^{xy} + 3*\lambda*y^2 = 0$$ $$(3)x^3+y^3 = 16$$

If $x=0$ or $y=0$ $==>$ $y=0$ or $x=0$ --> false if $t=0$ $==>$ $x=y=0$ ---> false

So, $x$,$y$ and $t$ cannot equal to 0.

So, we have from (1) (2) and (3): $$\frac{e^{xy}}{-3*\lambda} = \frac{x^2}{y}$$ $$\frac{e^{xy}}{-3*\lambda} = \frac{y^2}{x}$$ $$==> x= y $$ ==>$$ x = y = 2 $$

That is my solution. I just have one no, so I cannot find both min and max. Maybe something wrong with my solution. Please helps me.

Thanks :)

  • $\begingroup$ Please avoid using multiline equations in the title in the future; it screws up the main page layout. $\endgroup$ May 20, 2012 at 10:41
  • $\begingroup$ @N.I ah, I understand. Because I just view fomular from other post , I don't really know different $$ and $ :D $\endgroup$
    – hqt
    May 20, 2012 at 10:43
  • $\begingroup$ Basically \$ is for small, inline equations and \$\$ is for big, centered equations $\endgroup$ May 20, 2012 at 10:44

2 Answers 2


You may not find more than one critical value when solving problems like this. In your case, as seems to be confirmed when this question was asked on this site, you only have one critical value. To see if it's a min or max, you need to do some more work, though (see the last post in that thread).

I think you can convince yourself why this makes sense though (and why you've found a max and not a min). For your constraint, we can take $x$ positive and as big as we like. This puts a unique condition on $y$ and in particular forces it to be very close to $x$ in magnitude but $y < 0$. Notice then that when you plug these numbers into $f(x,y)$, you get ("large" means large absolute magnitude) $$f(x,y) = e^{xy} = e^{(\text{large positive)} \cdot (\text{large negative})} = e^{\text{large negative}} \approx 0.$$

But notice we can't ever get to $0$, only arbitrarily close. This is why there is no other critical point corresponding to a relative minimum: there is no relative minimum.


As of maximum value of $f(x,y)$

$$\frac{x^3+y^3}{2} \geq {\left(\frac{x+y}{2} \right)}^3 \geq {\left(\frac{2\sqrt{xy}}{2} \right)}^3 \geq ({xy})^\frac{3}{2}$$

Using the identity :

$$\frac{a_1^m+a_2^m+a_3^m...a_n^m}{n} \geq {\left(\frac{a_1+a_2+a_3...+a_n}{n} \right)}^m$$

and $A.M \geq G.M$

we get,

$$ \:\;\;\;\;\;\;\;\;\;\ xy\;\;\;\;\leq\;\;\;\;\; 4$$

$$\longrightarrow \:\;\;\;\;\;\;\;\;\;\ e^{xy}\;\;\;\;\leq\;\;\;\;\; e^4$$

  • $\begingroup$ Oh. It likes x=y=2 case when I use Lagrange Multiplier. But at min case, I don't have any idea how to solve by Lagrange Multiplier $\endgroup$
    – hqt
    May 20, 2012 at 11:03
  • 1
    $\begingroup$ Yes I know that , I wrote this answer so that you know that x=y=2 by lagrange multiplier is correct for maximum case as you have cited in your question, as of the minimum case I am thinking on it. :) $\endgroup$ May 20, 2012 at 11:16

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