To find the maximum profit, we need to follow these steps systematically:
### Step 1: Define the Profit Function
The profit function [tex]\( P(x) \)[/tex] is given by the difference between the revenue function [tex]\( R(x) \)[/tex] and the cost function [tex]\( C(x) \)[/tex]:
[tex]\[ P(x) = R(x) - C(x) \][/tex]
### Step 2: Express the Revenue and Cost Functions
The revenue function [tex]\( R(x) \)[/tex] is:
[tex]\[ R(x) = 120x - 0.015x^2 \][/tex]
The cost function [tex]\( C(x) \)[/tex] is:
[tex]\[ C(x) = 9000 + 60x - 0.03x^2 + 0.00001x^3 \][/tex]
### Step 3: Formulate the Profit Function
Substituting the revenue and cost functions into the profit function, we get:
[tex]\[
P(x) = (120x - 0.015x^2) - (9000 + 60x - 0.03x^2 + 0.00001x^3)
\][/tex]
Simplifying this, we obtain:
[tex]\[
P(x) = 120x - 0.015x^2 - 9000 - 60x + 0.03x^2 - 0.00001x^3
\][/tex]
[tex]\[
P(x) = 60x + 0.015x^2 - 0.00001x^3 - 9000
\][/tex]
### Step 4: Find the First Derivative of the Profit Function
To find the maximum profit, we need to take the first derivative of [tex]\( P(x) \)[/tex] and set it equal to zero:
[tex]\[
P'(x) = \frac{d}{dx} \left(60x + 0.015x^2 - 0.00001x^3 - 9000\right)
\][/tex]
[tex]\[
P'(x) = 60 + 0.03x - 0.00003x^2
\][/tex]
### Step 5: Solve for the Critical Points
We set the first derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
60 + 0.03x - 0.00003x^2 = 0
\][/tex]
This is a quadratic equation in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[
-0.00003x^2 + 0.03x + 60 = 0
\][/tex]
We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = -0.00003 \)[/tex], [tex]\( b = 0.03 \)[/tex], and [tex]\( c = 60 \)[/tex].
[tex]\[
x = \frac{-0.03 \pm \sqrt{(0.03)^2 - 4(-0.00003)(60)}}{2(-0.00003)}
\][/tex]
[tex]\[
x = \frac{-0.03 \pm \sqrt{0.0009 + 0.0072}}{-0.00006}
\][/tex]
[tex]\[
x = \frac{-0.03 \pm \sqrt{0.0081}}{-0.00006}
\][/tex]
[tex]\[
x = \frac{-0.03 \pm 0.09}{-0.00006}
\][/tex]
This gives us two solutions:
[tex]\[
x_1 = \frac{-0.03 + 0.09}{-0.00006} = \frac{0.06}{-0.00006} = -1000 \quad (\text{Not valid as we cannot produce a negative number of units})
\][/tex]
[tex]\[
x_2 = \frac{-0.03 - 0.09}{-0.00006} = \frac{-0.12}{-0.00006} = 2000
\][/tex]
So, the number of units that will give the maximum profit is [tex]\( \boxed{2000} \)[/tex].
### Step 6: Calculate the Maximum Possible Profit
Now, we substitute [tex]\( x = 2000 \)[/tex] back into the profit function [tex]\( P(x) \)[/tex]:
[tex]\[
R(2000) = 120(2000) - 0.015(2000^2) = 240000 - 60000 = 180000
\][/tex]
[tex]\[
C(2000) = 9000 + 60(2000) - 0.03(2000^2) + 0.00001(2000^3)
\][/tex]
[tex]\[
C(2000) = 9000 + 120000 - 120000 + 8000 = 137000
\][/tex]
[tex]\[
P(2000) = R(2000) - C(2000) = 180000 - 137000 = 43000
\][/tex]
The maximum possible profit is [tex]\( \boxed{43000} \)[/tex].
