A manufacturing unit currently operates at 80 percent of its capacity.

Question

A manufacturing unit currently operates at 80 percent of its capacity. The profit function for the unit at the optimum output, x, is given by p(x) = -0.1x2 + 80x − 60.

If the function f(x) models the current capacity of the unit, the composite function giving the unit's current profit function is

A). f(p(x)) = -0.064x^2 + 6.4x - 60

B). (f(x)) = -0.64x^2 + 0.64x - 60

C). p(f(x)) = -0.064x^2 + 64x - 60

D.) p(f(x)) = -0.64x^2 + 6.4x - 60

E.) f(p(x)) = -0.064x^2 + 64x - 60 .



If the optimum output is 500 units, the current profit is

A.)$ 15,400

B.) 15,940

C).16,060

D.)16,600

 

Ive been stuck here for a while

Answer 1

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)).

Answer 2

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)