# Two-Variable Optimisation Confusion

1. May 6, 2013

### JoshMaths

Hi,

So f(x,y) = xe-x(y2 - 4y)
Find all stationary points and classify them i got

for fx(x,y) s.p (1,4),(1,0)
for fy (x,y) s.p (0,2)

I thought that you don't need double differentials at this stage and if it is a s.p it must satisfy

for fx(x0,y0) = 0
for fy (x0,y0) = 0

which means s.p must hold for both partial derivatives?

I know the s.ps are wrong so if someone could advise that would be great.

Josh

2. May 6, 2013

### D H

Staff Emeritus
For a multivariate function, a point is a stationary point (better: a critical point) if all of the partial derivatives are zero. In your case, you need fx(x,y) and fy(x,y) to be zero for a point (x,y) to be a critical point. You went wrong in another regard as well. Consider $\frac{\partial f(x,y)}{\partial y} = xe^{-x}(2y-4)$. You have (0,2) as the only point at which fy(x,y)=0. That's not correct. fy(x,y) is a product. A product is zero if any of its factors is zero. Thus fy(x,y) is zero whenever x is 0 or when y is 2.

3. May 10, 2013

### JoshMaths

Ah so for fy(x,y) = 0 the solutions would be (0,y) and (x,2) as you can have any value for x and y for respective values that make the partial derivative zero?