Sure, I'll walk you through how we can determine the number of units that will give the maximum profit and the corresponding maximum possible profit.
### Step-by-Step Solution
1. Define the Revenue and Cost Functions:
- The revenue function [tex]\( R(x) \)[/tex] is given by [tex]\( R(x) = 205.2x - 0.03x^2 \)[/tex].
- The cost function [tex]\( C(x) \)[/tex] is given by [tex]\( C(x) = 9000 + 102.6x - 0.06x^2 + 0.00002x^3 \)[/tex].
2. Define the Profit Function:
- The profit function [tex]\( P(x) \)[/tex] is the revenue function minus the cost function:
[tex]\[
P(x) = R(x) - C(x) = (205.2x - 0.03x^2) - (9000 + 102.6x - 0.06x^2 + 0.00002x^3)
\][/tex]
Simplifying this, we get:
[tex]\[
P(x) = 205.2x - 0.03x^2 - 9000 - 102.6x + 0.06x^2 - 0.00002x^3
\][/tex]
[tex]\[
P(x) = (205.2x - 102.6x) + (-0.03x^2 + 0.06x^2) - 0.00002x^3 - 9000
\][/tex]
[tex]\[
P(x) = 102.6x + 0.03x^2 - 0.00002x^3 - 9000
\][/tex]
3. Find the Derivative of the Profit Function:
- Differentiating [tex]\( P(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[
P'(x) = \frac{d}{dx}[102.6x + 0.03x^2 - 0.00002x^3 - 9000]
\][/tex]
[tex]\[
P'(x) = 102.6 + 0.06x - 0.00006x^2
\][/tex]
4. Solve for Critical Points:
- To find the critical points, set the derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
102.6 + 0.06x - 0.00006x^2 = 0
\][/tex]
Solving this quadratic equation will give us the critical points.
5. Evaluate the Profit Function at the Critical Points:
- Substitute the critical points back into [tex]\( P(x) \)[/tex] to evaluate the profit at those points.
6. Determine the Maximum Profit:
- Among the evaluated profit values, find the maximum profit value. The corresponding [tex]\( x \)[/tex]-value will be the number of units that give the maximum profit.
After performing these calculations (and rounding to the nearest whole number if necessary), we find that:
(a) The number of units that will give the maximum profit is [tex]\( \boxed{1900} \)[/tex] units.
(b) The maximum possible profit is \$ [tex]\( \boxed{157060} \)[/tex].
