# Maxima/Minima in a conceptual equation

1. Sep 19, 2011

### elfy

1. The problem statement, all variables and given/known data
Find the stationary points of the function:
z = ax^2 + by^2 + c
For each of the following sub-cases, identify any maxima and minima.
i) a > 0, b > 0
ii) a < 0, b < 0
iii) a and b of opposite signs.

2. Relevant equations
z = ax^2 + by^2 + c

3. The attempt at a solution
Z'(x) = 2ax = 0
Z'(y) = 2by = 0

Z'(x) = Z'(y) => 2ax = 2by
Solving for y: Y = ax/b
Solving for x: X = by/a

Z''(xx) = 2a
Z''(yy) = 2b
Z''(xy) = 0

Checking the condition [Z''(xx) * Z''(yy)] - (Z''(xy))^2 --> (2a*2b) - (0)^2 = 4ab.

For i) where a and b are > 0 --> 4ab > 0 and 2a > 0 ---> Minimum
For ii) where a and b are < 0 --> 4ab > 0 and 2a < 0 ---> Maximum
For iii) where a and b are opposite --> 4ab < 0 --> Saddle Point

Have I done this correctly? Because I can help thinking that I have forgotten something. I would really appriciate any help and input with regards to my attempted solution! - Thanks!

2. Sep 19, 2011

### dynamicsolo

This looks to be correct (though your partial derivative notation should be something more like zxxzyy - ( zxy )2 , for instance).

This surface is an easy one to check up on. If you alternately set either x = 0 or y = 0 , you find that the cross-sections on the yz- and xz-planes are parabolas. When the signs of a and b match, both parabolic cross-sections open in the same direction. When the signs are opposite, the parabolas are opening in opposite directions.

So having both a and b positive gives us an (elliptic) paraboloid which opens upward (in the positive-z direction), meaning that the critical point at (x, y) = (0, 0) is a minimum. Both a and b being negative has the paraboloid opening downward, so (0, 0) becomes a maximum. For a and b having opposite signs, the surface is a hyperbolic paraboloid and (0, 0) is the saddle point.

3. Sep 19, 2011

### elfy

Hey dynami, and thanks for your help!

I'm afraid I am not as confident as you are when it comes to shapes in the xy plane so I dont think I have a chance of "seeing" what you just described without actually plotting the graphs manually on a piece of paper hehe.

However, what you just described, does that always hold? I.e. if a and b are both positive, the paraboloid will always open upwards? What I mean is, can I apply that reasoning to any equation (with actual numbers replacing a and b) and "see" the form straight away before even doing the check with the double partial derivatives?