1
$\begingroup$

I am having trouble understanding how to solve the problem below. Can anyone show me how to solve this? Here is the problem definition:

"Use LaGrange multipliers to find the maximum and minimum values of the function $f(x,y)=e^{xy}$ subject to the constraint $x^3+y^3=16$."
This is problem number 14.8.6 in the seventh edition of Stewart Calculus.

My work so far:
Use $\nabla f=\lambda\nabla g$ and $g(x,y)=16$
$f_x=ye^{xy}=\lambda g_x=\lambda 3x^2$
$f_y=xe^{xy}=\lambda g_y=\lambda 3y^2$
Solving these equations gives me $x=1, y=1, \lambda=\frac{e}{3}$
However, I am confused because $f(1,1)=e$, but $g(1,1)=2\ne 16$
How do I finish this problem correctly?

$\endgroup$
  • $\begingroup$ I think you might have an error in your calculations; I found $x=y=2$ and $\lambda=e^4/6$. $\endgroup$ – user12477 Oct 30 '12 at 22:48
  • 1
    $\begingroup$ @user12477 Thank you. I see how $x=y=2$ and $\lambda = \frac{e^4}{6}$ plug into the equation, but I do not see how to show the work to get those numbers without simply guessing them. Also, that is just one point. I have to find both maximum and minimum. Do you have any further suggestions? $\endgroup$ – CodeMed Oct 30 '12 at 22:55
3
$\begingroup$

Start with the equations that you have derived: \begin{eqnarray*} ye^{xy}&=&3\lambda x^2,\\ xe^{xy}&=&3\lambda y^2,\\ x^3+y^3&=&16. \end{eqnarray*} As a first step, show that none of $\lambda, x$ or $y$ can be zero. (If one of them is zero, then the first two equations show that all three must be zero, contradicting the third equation.) This is useful as we now know that we can divide by these terms at will.

Now comparing the first and second equations gives $x^3=y^3$, and since both are real, we get $x=y$. The third equation then gives $x=y=2$, and the first or second yields the value of $\lambda$. We find $f=e^4$ at this point, and it must be a maximum.

As regards the minimum, recall that the Lagrange multiplier method identifies the possible location of max/min points IF they exist. I don't think your example has a minimum: By taking $x^3 = N^3$ and $y^3=16-N^3$, where $N$ is a large positive number, the constraint $x^3+y^3=16$ is satisfied. But $xy\sim-N^2$ can be made arbitrarily large and negative, so that $e^{xy}$ can be made as close as we like to $0$, but of course would never equal zero. So we can say that the infimum $$\inf\{e^{xy}:x^3+y^3=16\}=0,$$ but the minimum $$\min\{e^{xy}:x^3+y^3=16\}$$ does not exist: for any candidate minimum at $(x_0,y_0)$, we can always find $(x_1,y_1)$ with $0<f(x_1,y_1)<f(x_0,y_0)$.


Another approach is to eliminate $y$ and to treat this as a single variable max/min problem for $h(x)=\exp[x(16-x^3)^{1/3}]$. This gives another way of understanding why there is no minimum point of the function.

$\endgroup$
  • $\begingroup$ Thank you very much. You state that $f(2,2)=e^4$ must be a maximum. Can you explain why? I think I understand the rest of your logic. +1 for helping me out. $\endgroup$ – CodeMed Oct 30 '12 at 23:42
  • $\begingroup$ It's certainly a local max from the Lagrange multiplier approach: the single variable approach shows that it is a global max. $\endgroup$ – user12477 Oct 30 '12 at 23:53

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.