Answer:
If the function f(x) models the current capacity of the unit, the composite function giving the unit's current profit function is
f(p(x)) = -0.064x^2 + 64x - 60. If the optimum output is 500 units, the current profit is $16,060.
Explanation:
Profit a max capacity is:
p(x) = -0.1x2 + 80x - 60
Current capacity, f(x), is 80% of max capacity:
f(x) = 0.8x
Profit at current capacity:
p(f(x)) = -0.1f(x)2 + 80f(x) - 60
p(f(x)) = -0.1(0.8x)2 + 80(0.8x) - 60
p(f(x)) = -0.064x2 + 64x - 60
To find the profit a current capacity, plug x = 500 into the above equation (p(f(x)).
Using the 80% equation I found, you would plug in 500 for x. Or, you
could replace x with 400 in the original p(x).
p(x) = -0.1x2 + 80x - 60
Since we are operating at 80% capacity,
replace x with .80x or .8x
p((fx)) = -.1(.8x)2 + 80(.8x) - 60
= -.064x2 + 64x - 60
This would be answer c
Optimum output is 500 units.
80% of 500 = .8(500) = 400 units output
In your original equation replace x with
400 and solve for p(x)