SOLUTION (Polymer production problem, leading to system of first order, linear discrete dynamical systems, and optimization):

We enter the matrix of coefficients for the transition from Material A to Material B and back again.

We compute the eigenvalues and the corresponding eigenvectors.

We write the solution in terms of the eigenvalues, the eigenvectors, and the constants c1 and c2. (The s[[i]][[j]] terms go to the ith list in s and take the jth value offered, so for example s[[2]][[1]] is 0.894427.)

We set our initial conditions with the unkown x.

We solve for the constants c1 and c2 - obtaining them in terms of the variable x. This part will not permit just plugging and chugging to get the coefficients, indeed, the inverse of the coefficient matrix is probably best, but then it has to be multiplied by the initial vector (1 + x, 1 - x^2).

We write out our amounts as a vector h(n) - being the amount of (Material A, Material B) at time n in terms of the variable x.

We compute the amount of material A and matrial B at time n = 10, still as a function of x.

We plot both - the thin line is Material A and the thick line is material B.

We determine the maximum value of the SECOND coordinate function (material B) by zooming in around x = 0.5.

We determine the maximum value of the SECOND coordinate function (material B) by zooming in around x = 0.46.

We determine the maximum value of the SECOND coordinate function (material B) by zooming in around x = 0.46.

And hence we get a maximum value of about 0.7506 g of B when we originally add

x = 0.46 g of A to our initial amount of 1 g of A and remove x^2 = 0.46^2 = .2116 g of B from our initial amount of 1 g of B.

Or we could determine the maximum value of the SECOND coordinate function (material B) by differentiating w.r.t. x and setting the derivative equal to 0 to solve for x.

And finally we determine the actual maximum amount of material B at time n = 10 when we take x = .4599 g more than 1 g of A as initial condition.

We plot the amount of material B in the case x = .4599.

We see that Material B is headed for 0.7495 g as a long term amount.