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I'm a TA for an honors calculus course and I've come upon a homework problem (for which I'm supposed to write solutions!) that has me stumped. This is problem 12, in Chapter 9.13 in volume 2 of Apostol's Calculus book.

Find the critical values of $f(x,y) = (5x+7y-25)e^{-(x^2 + xy + y^2)}$ and classify them as max/min/saddle points.

The back of the book gives the solution as absolute min at $(-\frac{1}{26}, -\frac{3}{26})$ and absolute max at $(1,3)$, so I'm not so interested in just knowing the points.

However, if one sets $f_x = 0 = f_y$ and solves simultaneously, one finds that, say, $y$ satisfies a quartic polynomial (and it's nothing simple like a quadratic polynomial in $y^2$). Now, I'm aware that we do know the quartic formula, but I'm thinking it shouldn't be necessary to use it.

In the paragraph before these exercises (this is problem 12), Apostol says (paraphrased) "When $f(x,y) = e^{1/g(x,y)}$ with $g(x,y) = x^2 + 2 + \cos^2 y -2\cos y$, we can find the critical points via the usual method, but if instead we write $g(x,y) = 1 + x^2 + (1-\cos y)^2$, we see that $f$ has a relative maxima at the points at which $x^2 = 0$ and $(1-\cos y)^2 = 0$".

I mention this because I think this particular exercise should be solved via a clever trick as in the previous paragraph, but I've not been able to come up with the trick on my own.

Is there a nice trick for finding the critical points?

Thank you!

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Would the question downvoter care to elaborate? – Jason DeVito Jun 16 '11 at 13:25
up vote 7 down vote accepted

Taking the logarithmic derivative simplifies the problem.

We obtain $$\frac{f_x}{f} = \partial_x \ln f = -2x -y + \frac{5}{5x +7y -25} = \frac{-(2x+y)(5x +7y -25)+5}{5x +7y -25}$$ and $$\frac{f_y}{f} = \partial_y \ln f = -2y -x + \frac{7}{5x +7y -25}=\frac{-(2y+x)(5x +7y -25)+7}{5x +7y -25}.$$

Both of them have to be zero at the critical point (unless $f =0$ which one should in principle check additionally). We obtain the set of equations $$-(2x+y)(5x +7y -25)+5=0$$ $$-(2y+x)(5x +7y -25)+7=0,$$ which can be solved easily.

Edit: (to show how the set of equations can be solved)

Multiplying the first equation with $2y+x$, the second equation with $2x+y$ and subtracting the results from each other yields $3y - 9x =0$, with the solution $y=3 x$. Plugging this result in the first equation, we obtain $-5x(26x -25) +5 =0$ with the solutions $x=1$ and $x=-1/26$.

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These are precisely the equations I had (maybe I should edit that in) which I couldn't "solve easily" without invoking the quartic formula. That is, if you, say, solve the first equation for $x$ and plug it into the second, and multiply out, you get a quartic in $y$. Maybe I'm just missing something stupid - would you mind editing and solving the set of equations? – Jason DeVito Mar 29 '11 at 23:28
@Janson DeVito: ok, I will add some lines... – Fabian Mar 29 '11 at 23:32
Beautiful - that's a nice trick!. I'm going to hold off on accepting this to see if anyone else has a nice trick - but this is definitely simple enough that I should have seen it! – Jason DeVito Mar 29 '11 at 23:44

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