T = 3A^2 + 100A+400 + 6B^2+150B+600
but B=500-A, so substituting this in the formula for T we have:
T = 9 A^2 - 6050 A + 1576000.
To find the lowest cost we differentiate T with respect to A and set the
derivative equal to zero and solve for A.
d/dA T = 6 A - 6050,
so for the minimum: A=6050/6, and B=500 - 6050/6. This gives the point
of global minimum for T (since T is a quadratic in A opening upwards this
stationary point is of necessity an global minimum rather than any other
king of stationary point).
Now I assume that we have to have A,B >=0, so this minimum is not
in the feasible region for this problem, so with these additional constraints
we know that the minimum lies on the boundard of the feasible region.
In this case the feasible region is the line segment from (0,500) to (500,0)
in the AB plane, so the boundary of the feasible region is the pair of points
(500,0) anf (0,500).
That is either A=0 or B=0, When B=0 T=801000, when A=0 T =2026000, so
the minimum occurs when A=500 and B=0, and the total cost is 801000.