Urban Community College is planning to offer courses in Finite Math, Applied Calculus, and Computer Methods. Each section of Finite Math has 40 students and earns the college $40,000 in revenue. Each section of Applied Calculus has 40 students and earns the college $60,000, while each section of Computer Methods has 10 students and earns the college $26,000. Assuming the college wishes to offer a total of six sections, accommodate 210 students, and bring in $266,000 in revenues, how many sections of each course should it offer?
Finite Math
section(s)
Applied Calculus
section(s)
Computer Methods
section(s)

To determine how many sections of each course Urban Community College should offer, let's set up the problem with variables:

Let:
- $ x $ = number of sections of Finite Math
- $ y $ = number of sections of Applied Calculus
- $ z $ = number of sections of Computer Methods

We have the following constraints based on the problem:

1. Total sections constraint:
   \[
   x + y + z = 6
   \]

2. Total students constraint:
   Each Finite Math section has 40 students, each Applied Calculus section has 40 students, and each Computer Methods section has 10 students:
   \[
   40x + 40y + 10z = 210
   \]

3. Total revenue constraint:
   The revenues from each course are $40,000 for Finite Math, $60,000 for Applied Calculus, and $26,000 for Computer Methods:
   \[
   40000x + 60000y + 26000z = 266000
   \]

Now we have a system of three equations:

1. $ x + y + z = 6 $
2. $ 40x + 40y + 10z = 210 $
3. $ 40000x + 60000y + 26000z = 266000 $

### Step 1: Simplify the equations

From equation 2, we can simplify by dividing everything by 10:
\[
4x + 4y + z = 21 \quad \text{(Equation 2')}
\]

Now we have:
1. $ x + y + z = 6 $ (Equation 1)
2. $ 4x + 4y + z = 21 $ (Equation 2')
3. $ 40000x + 60000y + 26000z = 266000 $ (Equation 3)

### Step 2: Solve for $ z $

We can express $ z $ from Equation 1:
\[
z = 6 - x - y
\]

### Step 3: Substitute $ z $ into Equation 2'

Substituting $ z $ into Equation 2':
\[
4x + 4y + (6 - x - y) = 21
\]
\[
4x + 4y + 6 - x - y = 21
\]
\[
3x + 3y + 6 = 21
\]
\[
3x + 3y = 15
\]
\[
x + y = 5 \quad \text{(Equation 4)}
\]

### Step 4: Substitute $ z $ into Equation 3

Substituting $ z $ into Equation 3:
\[
40000x + 60000y + 26000(6 - x - y) = 266000
\]
Expanding:
\[
40000x + 60000y + 156000 - 26000x - 26000y = 266000
\]
Combining like terms:
\[
(40000 - 26000)x + (60000 - 26000)y + 156000 = 266000
\]
\[
14000x + 34000y + 156000 = 266000
\]
\[
14000x + 34000y = 110000 \quad \text{(Equation 5)}
\]

### Step 5: Solve Equations 4 and 5

From Equation 4:
\[
y = 5 - x
\]

Substituting into Equation 5:
\[
14000x + 34000(5 - x) = 110000
\]
Expanding:
\[
14000x + 170000 - 34000x = 110000
\]
Combining like terms:
\[
-20000x + 170000 = 110000
\]
\[
-20000x = -60000
\]
\[
x = 3
\]

### Step 6: Find $ y $ and $ z $

Substituting $ x $ back into Equation 4:
\[
3 + y = 5 \implies y = 2
\]

Now substituting $ x $ and $ y $ back into Equation 1 to find $ z $:
\[
3 + 2 + z = 6 \implies z = 1
\]

### Final Solution

- **Sections of Finite Math:** $ x = 3 $
- **Sections of Applied Calculus:** $ y = 2 $
- **Sections of Computer Methods:** $ z = 1 $

### Conclusion

Urban Community College should offer:
- **Finite Math:** 3 sections
- **Applied Calculus:** 2 sections
- **Computer Methods:** 1 section

Would you like further details or have any questions? Here are some related questions you might consider:

1. How would changing the number of sections affect total revenue?
2. What if the college wanted to accommodate more students?
3. How do the costs of offering each course compare?
4. What are the potential enrollment numbers if class sizes were adjusted?
5. How can the college better meet student demand with the current offerings?

**Tip:** When solving similar problems, always define your variables clearly and set up equations based on given constraints.

Urban Community College is planning to offer courses in Finite Math, Applied Calculus, and Computer Methods. Each section of Finite Math has 40 students and earns the college $40,000 in revenue. Each section of Applied Calculus has 40 students and earns the college $60,000, while each section of Computer Methods has 10 students and earns the college $26,000. Assuming the college wishes to offer a total of six sections, accommodate 210 students, and bring in $266,000 in revenues, how many sections of each course should it offer?

Determine the optimal number of sections for Finite Math, Applied Calculus, and Computer Methods at Urban Community College to meet constraints on total sections, student capacity, and revenue. This problem demonstrates the use of systems of linear equations and algebraic methods, ideal for advanced high school or introductory college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation