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 $.

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

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OrethaWilkison OrethaWilkison · Answer 2 · 2019-05-13T09:40:21+02:00

Answer:

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

Step-by-step explanation:

As per the statement:

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

[tex]p(x) =-0.1x^2+ 80x -60[/tex]

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:

[tex]p(f(x)) = p(0.80x)[/tex]

⇒[tex]p(0.80x) = -0.1(0.80x)^2+80(0.80x)-60[/tex]

Simplify:-

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

We have to find the current profit.

If the optimum output is 500 units

⇒x = 500 units

Substitute in [1] we have;

[tex]p(f(500)) = -0.064(500)^2+64(500)-60[/tex]

⇒[tex]p(f(500)) = -16000+32000-60[/tex]

Simplify:

[tex]p(f(500)) =\$ 15,940[/tex]

Therefore, the current profit is $15,940