
Maximizing Hard Lin Alg
Hi,
I am trying to find a rho that maximizes r..
The orig func. is r^4  ar^3 + br^2 + r*R_(0)/2*(a  2) + R_(0)/2*(a  2b) = 0
Then,
4r^3(dr/dp)  3ar^2(dr/dp) + 2br(dr/dp) + R_(0)/2*(a  2)*(dr/dp) + (r/2)*(a  2)(dR_(0)/dp) + (1/2)*(dR_(0)/dp)*(a  2b) = 0
Note the dr/dp and dR_0/dp is the partial of r with respect to rho and the partial of R_(0) with respect to rho.
And, once I have found out the rho that maximizes r, how can I find a value of R_(0) that maximizes rho?
So I think that's just (dR_(0)/dp) = 0 and then determine hoow R_(0) changes with rho.

1) Please retype in Latex  it's much easier to read that way.
2) Please verify that you read the problem correctly. As it stands, you can make r as large as you want by choosing R_(0) sufficiently negative.