# Finding the stationary point of the following function

#### u-will-neva-no

Joined Mar 22, 2011
230
Hey everyone, my problem is finding the stationary point of the function:

$$f(x,y) = (x+y)exp(x^2-y)$$

By differentiating f WRT x i obtained the answer:
$$\frac{df}{dx} = exp(x^2-y)[2x^2+2xy+1] = 0$$ (1)

and differentiating f WRT y I got:

$$\frac{df}{dy} = exp(x^2-y)[1-x-y]$$ (2)

I could be wrong with my answer because I am now stuck because I always have one equation that is a function of two variables (x and y). However, I have tried three times and arrive at the same solution for both. Please help!

Thanks!!

#### steveb

Joined Jul 3, 2008
2,436
You have two equations and two unknowns. So, can't you just solve these simultaneous equations to find any stationary points?

• u-will-neva-no

#### Papabravo

Joined Feb 24, 2006
18,431
You have two equations and two unknowns. So, can't you just solve these simultaneous equations to find any stationary points?
Yes except the equations are non-linear. Techniques for solving simultaneous non-linear equations not that straight forward. I would try a 3-D graph to see if the surfaces intersect. One equation is a plane and the other is a conic solid, so i'm guessing the intersection is a conic section. A unique single point solution would be a surprise.

• u-will-neva-no

#### steveb

Joined Jul 3, 2008
2,436
A unique single point solution would be a surprise.
Prepare to be surprised. The exponential parts can be divided out of the equations because they do not cause the equations to equal zero, since exponential is always a positive nonzero number (assuming real arguments). Only the polynomials in x and y are left, once that is done, and those can be solved very easily in this case.

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• u-will-neva-no

#### Papabravo

Joined Feb 24, 2006
18,431
Prepare to be surprised. The exponential parts can be divided out of the equations because they do not cause the equations to equal zero, since exponential is always a positive nonzero number (assuming real arguments). Only the polynomials in x and y are left, once that is done, and those can be solved very easily in this case.
I knew the exponential parts could not be zero, but I did expect the intersection of a plane and a conic section to be more than a single point.

• u-will-neva-no

#### steveb

Joined Jul 3, 2008
2,436
... but I did expect the intersection of a plane and a conic section to be more than a single point.
Ah, but that is not quite what it is that gives the answer. The answer is not just the intersection, but only points on the intersection that allow both independent equations to equal zero in the following sense.

One equation can be written as f(x,y)=0 and the other can be written as g(x,y)=0 and they must both be zero at the stationary point. The intersection is more general and amounts to the equation f(x,y)=g(x,y) which is not the same thing.

One also has to consider the context of the original question, which is to find the stationary point. So, the question is telling you that there is a stationary point. Perhaps there will be more than one point in general, but typically sufaces are stationary (minimum, maximum or saddle) at a single point locally. There are exceptions of course, but these are exceptions to the rule.

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• u-will-neva-no