Question

The weekly revenue for a product is given by ​R(x)equals288 x minus 0.054 x squared​, and the weekly cost is ​C(x)equals10 comma 000 plus 144 x minus 0.108 x squared plus 0.00003 x cubed​, where x is the number of units produced and sold.
​(a) How many units will give the maximum​ profit?
​(b) What is the maximum possible​ profit?

Answer 1

To find the maximum profit, we first need to determine the profit function \(P(x)\), which is defined as the difference between the revenue function \(R(x)\) and the cost function \(C(x)\).

### Given:
- Revenue:
\[
R(x) = 288x - 0.054x^2
\]

- Cost:
\[
C(x) = 10,000 + 144x - 0.108x^2 + 0.00003x^3
\]

### Profit Function:
The profit function \(P(x)\) is given by:
\[
P(x) = R(x) - C(x)
\]
Substituting in the expressions for \(R(x)\) and \(C(x)\):
\[
P(x) = (288x - 0.054x^2) - (10,000 + 144x - 0.108x^2 + 0.00003x^3)
\]

Now, let's simplify this:
\[
P(x) = 288x - 0.054x^2 - 10,000 - 144x + 0.108x^2 - 0.00003x^3
\]
Combining like terms:
\[
P(x) = (288 - 144)x + (-0.054 + 0.108)x^2 - 0.00003x^3 - 10,000
\]
\[
P(x) = 144x + 0.054x^2 - 0.00003x^3 - 10,000
\]

### (a) Finding the Maximum Profit:
To find the maximum profit, we need to find the critical points by taking the derivative of \(P(x)\) and setting it equal to zero.
\[
P'(x) = 144 + 0.108x - 0.00009x^2
\]
Setting \(P'(x) = 0\):
\[
144 + 0.108x - 0.00009x^2 = 0
\]
Rearranging gives:
\[
0.00009x^2 - 0.108x - 144 = 0
\]

Next, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 0.00009\), \(b = -0.108\), and \(c = -144\).

Calculating the discriminant:
\[
b^2 - 4ac = (-0.108)^2 - 4(0.00009)(-144)
\]
\[
= 0.011664 + 0.05184 = 0.063504
\]

Now we can find \(x\):
\[
x = \frac{-(-0.108) \pm \sqrt{0.063504}}{2(0.00009)}
\]
Calculating \( \sqrt{0.063504} \):
\[
\sqrt{0.063504} \approx 0.252
\]
\[
x = \frac{0.108 \pm 0.252}{0.00018}
\]
Calculating both potential values:
\[
x_1 = \frac{0.360}{0.00018} \approx 2000
\]
\[
x_2 = \frac{-0.144}{0.00018} \text{ (not relevant since } x \text{ cannot be negative)}
\]

Thus, the number of units that maximizes profit is:
\[
\boxed{2000}
\]

### (b) Calculating the Maximum Profit:
Now substituting \(x = 2000\) into the profit function \(P(x)\):
\[
P(2000) = 144(2000) + 0.054(2000^2) - 0.00003(2000^3) - 10,000
\]
Calculating each term:
- \(144(2000) = 288000\)
- \(0.054(2000^2) = 0.054 \times 4000000 = 216000\)
- \(0.00003(2000^3) = 0.00003 \times 8000000000 = 240000\)

Now substitute these values back into the profit formula:
\[
P(2000) = 288000 + 216000 - 240000 - 10,000
\]
\[
= 288000 + 216000 - 240000 - 10,000 = 254000
\]

Thus, the maximum possible profit is:
\[
\boxed{254000}
\]