Question

The total revenue function for a product is given by Requals745x ​dollars, and the total cost function for this same product is given by Cequals10 comma 500plus30xplusxsquared​, where C is measured in dollars. For both​ functions, the input x is the number of units produced and sold.
a. Form the profit function for this product from the two given functions.
b. What is the profit when 13 units are produced and​ sold?
c. What is the profit when 21 units are produced and​ sold?
d. How many units must be sold to break even on this​ product?

Answer 1

To address the problem, we'll first derive the profit function based on the provided revenue and cost functions.

### Part (a): Form the Profit Function
The profit function \( P(x) \) is calculated as the difference between the total revenue \( R(x) \) and the total cost \( C(x) \):

1. **Total Revenue**:
\[ R(x) = 745x \]

2. **Total Cost**:
\[ C(x) = 10,500 + 30x + x^2 \]

3. **Profit Function**:
\[ P(x) = R(x) - C(x) \]
\[ P(x) = 745x - (10,500 + 30x + x^2) \]
\[ P(x) = 745x - 10,500 - 30x - x^2 \]
\[ P(x) = (745 - 30)x - 10,500 - x^2 \]
\[ P(x) = 715x - x^2 - 10,500 \]

So, the profit function is:
\[
P(x) = -x^2 + 715x - 10,500
\]

### Part (b): Profit when 13 Units are Produced and Sold
To find the profit at \( x = 13 \):
\[
P(13) = -13^2 + 715(13) - 10,500
\]
\[
= -169 + 9295 - 10,500
\]
\[
= -169 + 9295 - 10,500 = -169 + 9295 - 10,500 = -169 - 1205 = -1374
\]

So, the profit when 13 units are produced and sold is:
\[
P(13) = -1374 \text{ dollars}
\]

### Part (c): Profit when 21 Units are Produced and Sold
To find the profit at \( x = 21 \):
\[
P(21) = -21^2 + 715(21) - 10,500
\]
\[
= -441 + 14915 - 10,500
\]
\[
= -441 + 14915 - 10,500 = -441 + 3875 = 3434
\]

So, the profit when 21 units are produced and sold is:
\[
P(21) = 3434 \text{ dollars}
\]

### Part (d): Break-even Point
To find the break-even point, we need to set the profit function to zero and solve for \( x \):
\[
P(x) = 0
\]
\[
-x^2 + 715x - 10,500 = 0
\]

To solve this quadratic equation, we can rearrange it:
\[
x^2 - 715x + 10,500 = 0
\]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 1 \)
- \( b = -715 \)
- \( c = 10,500 \)

Now calculate the discriminant:
\[
b^2 - 4ac = (-715)^2 - 4(1)(10,500)
\]
\[
= 511225 - 42000 = 469225
\]

Now find the square root of the discriminant:
\[
\sqrt{469225} = 685
\]

Using the quadratic formula:
\[
x = \frac{715 \pm 685}{2}
\]

Calculate the two potential solutions:
1. \( x = \frac{715 + 685}{2} = \frac{1400}{2} = 700 \)
2. \( x = \frac{715 - 685}{2} = \frac{30}{2} = 15 \)

Thus, the break-even points are \( x = 15 \) and \( x = 700 \). However, for business analysis, we commonly refer to the first break-even point at reasonable production levels:
\[
x = 15
\]

So, the number of units that must be sold to break even on this product is:
\[
\text{Break-even units} = 15
\]

### Summary:
- **Profit Function**: \( P(x) = -x^2 + 715x - 10,500 \)
- **Profit for 13 units**: -1374 dollars
- **Profit for 21 units**: 3434 dollars
- **Break-even point**: 15 units