Thread: 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.