Answer:
2000 of Type A; 3000 of Type B; profit of $74 million
Explanation:
You have enough of these similar questions that it seems like it would pay to find a generic solution.
Let revenue (R) be ...
R = ax + by
and cost (C) be ...
C = cx² +dxy +ey² +fx +gy +h
Then profit (P) is the difference between revenue and cost ...
P = R - C = ax + by - (cx² +dxy +ey² +fx +gy +h)
P = -cx² -dxy -ey² +(a-f)x +(b-g)y -h
The maximum is found where the partial derivatives with respect to x and y are zero.
∂P/∂x = -2cx -dy +(a-f) = 0
∂P/∂y = -dx -2ey +(b-g) = 0
These two equations can be written in standard form as ...
2cx +dy = a-f
dx +2ey = b-g
Then Cramer's rule gives the generic solution to this linear system as ...
x = (d(b-g) -2e(a-f))/(d² -4ce)
y = (d(a-f) -2c(b-g)/(d² -4ce)
_____
Using your numbers ...
{a, b, c, d, e, f, g, h} = {4, 6, 1, -3, 9, 9, -42, -7}
we get ...
x = (-3(6-(-42)) -2·9(4-9))/((-3)² -4·1·9) = (-3·48 -18(-5))/-27 = -54/-27 = 2
y = (-3(-5) -2·1·48)/-27 = -81/-27 = 3
And the profit is ...
P = -1(2²) +3(2)(3) -9(3²) -5(2) +48(3) +7 = -4 +18 -81 -10 +144 +7 = 74
Sale of 2000 type A units and 3000 type B units will yield the maximum profit of $74 million.