To determine how rapidly the revenue is increasing when 380 units have been sold and sales are increasing at a rate of 55 units per day, we need to find the rate of change of the revenue with respect to time.
First, we are given the revenue function [tex]\( R \)[/tex] as a function of [tex]\( x \)[/tex]:
[tex]\[ R(x) = 2000x - 2x^2 \][/tex]
To find how the revenue is changing with respect to [tex]\( x \)[/tex], we compute the first derivative of [tex]\( R \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dR}{dx} = \frac{d}{dx}(2000x - 2x^2) \][/tex]
Using the power rule and linearity of differentiation, we have:
[tex]\[ \frac{dR}{dx} = 2000 - 4x \][/tex]
Next, we need to evaluate this derivative at [tex]\( x = 380 \)[/tex] units:
[tex]\[ \frac{dR}{dx} \bigg|_{x=380} = 2000 - 4(380) \][/tex]
[tex]\[ \frac{dR}{dx} \bigg|_{x=380} = 2000 - 1520 \][/tex]
[tex]\[ \frac{dR}{dx} \bigg|_{x=380} = 480 \][/tex]
This value, 480, represents the rate of change of revenue with respect to the number of units sold.
Since sales are increasing at the rate of 55 units per day, the revenue increase per day is given by the product of the rate of change of revenue with respect to units sold and the rate at which sales are increasing:
[tex]\[ \frac{dR}{dt} = \frac{dR}{dx} \cdot \frac{dx}{dt} = (480) \cdot (55) \][/tex]
Finally, calculate the revenue increase per day:
[tex]\[ \frac{dR}{dt} = 480 \times 55 \][/tex]
[tex]\[ \frac{dR}{dt} = 26400 \][/tex]
Thus, the revenue is increasing at a rate of:
[tex]\[ 26400 \text{ dollars per day} \][/tex]
So the answer is:
[tex]\[ \boxed{26400 \text{ dollars per day}} \][/tex]
