Expert: Sherry Wallin - 11/20/2009

Question
Dear Sherry, pls help me with the following problems.

1) How to find the extreme values and test if they are maxima or minima:
z=29-x^2+y^2+m^2

2)Following quadratic form is given. How to express it as a matrix product involving a symmetric coefficient matrix. How to determine if this matrix is positive definite, negative definite, or something else.
q=3a^2-2ab+4ac+5b^2+4c^2-2bc

Answer
for #1 What is m? Is it a variable or a constant? If it is a constant then take the first partial derivatives with respect to each variable, set those equations to zero find their critical values, see what the function is doing (increasing, decreasing etc). You need to find the extreme values of f on the boundary D (which you didn't give me D). Whatever D is, it is bounded by some function or lines and you need to check f at each of the boundaries and whichever (either the critical values or the boundary values are largest/smallest will be your max or min respectively.

#2 we will break this problem into parts and then put it back together
part I:  3a^2-2ab+4ac+5b^2 you need to complete the square using a as your variable and treating bc as constants:

3a^2-2(b-2c)a + 5b^2 = 3(a^2 -(2/3)(b-2c)a)+ 5b^2
= 3(a^2 -(2/3)(b-2c)a + [(-1/3)(b-2c)]^2 + 5b^2 -3[(-1/3)(b-2c)]^2
= 3(a-(1/3)(b-2c))^2 + 5b^2 -(1/9)(b^2-4bc+4c^2)
= 3(a-(1/3)(b-2c))^2 + 5b^2 -(1/3)b^2 + (4/3)bc -(4/3)c^2
= 3(a-(1/3)(b-2c))^2 + (14/3)b^2 + (4/3)bc -(4/3)c^2

Now is the time to add part II: 4c^2-2bc giving us:

3(a-(1/3)(b-2c))^2 + (14/3)b^2 + (4/3)bc -(4/3)c^2 + 4c^2-2bc but now focus on the just the parts with the b's and c's in them:

(14/3)b^2 + (4/3)bc -(4/3)c^2 + 4c^2 -2bc
= (14/3)b^2 -(2/3)bc + (8/3)c^2
= (14/3)(b^2-(1/7)bc ) + (8/3)c^2
= (14/3)(b^2-(1/7)bc + ((-1/14)c)^2) -(14/3)[(1/14)c]^2 + (8/3)c^2
= (14/3)(b^2-(1/14)c))^2 - (1/42)c^2 + (8/3)c^2
= (14/3)(b^2-(1/14)c))^2 + (111/4)c^2

Now you put all the pieces together and get:
q = 3(a-(1/3)(b-2c))^2 + (14/3)(b^2-(1/14)c))^2 + (111/42)c^2
q is now seen to be composed of the sum of 3 squares which will always be positive unless all of a,b,c are zero, meaning the zero vector, so this is positive definite.

Math Prof

I hope there aren't any arithmetical errors, there is a lot of room (places) to make them. One way to check it out is to expand what I have and see if it is the same as what you began with or to try small values for a,b,c in the original and the last and see if they generate the same answer. I used a=b=c=1 and both give me 12 for an answer so it is probably correct arithmetically.