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

Answer 1

If the optimum output is 500 units, the current profit is $ 15,940

Explanation:

As per the statement:

The profit function for the unit at the optimum output, x, is given by:

`p(x) =-0.1x^2+ 80x -60`

If the function f(x) models the current capacity of the unit.

⇒we substitute f(x) =0.80x for every x in the equation.

then;

the composite function giving the unit's current profit function is:

`p(f(x)) = p(0.80x)`

⇒
`p(0.80x) = -0.1(0.80x)^2+80(0.80x)-60`

Simplify:-

`p(f(x)) = -0.064x^2+64x-60` ....[1]

We have to find the current profit.

If the optimum output is 500 units

⇒x = 500 units

Substitute in [1] we have;

`p(f(500)) = -0.064(500)^2+64(500)-60`

⇒
`p(f(500)) = -16000+32000-60`

Simplify:

`p(f(500)) =\$ 15,940`

Therefore, the current profit is $15,940

Answer 2

The profit function is -0.1x² + 80x - 60

To account for 80% capacity, we substitute f(x) = .8x for every x in the equation:
p(f(x)) = -.1(.8x)^2 + 80(0.8x) - 60
p(f(x)) = -0.064x² + 64x - 60

Plugging in 500 for x:
p(f(500)) = -0.064 (500²) + 64 (500) -60
p(f(500)) = 15940