﻿A biscuit factory's variable cost per month in hundreds of dollars are given by the equation
q
-
1
2
0
,
 ﻿where 
q
 ﻿represents batches of biscuits. The company incurs a fixed cost of $
3
6
2
0
.
Determine the total cost function. How many biscuits need to be produced to minimize the
total cost. 
1
 ﻿batches of biscuits
Now determine the average cost function. Based on the batches of biscuits that minimizes the
total cost, what is the average cost for biscuits? 
(
round to the nearest dollar
)
 ﻿$


We are given:

- The variable cost function: $ VC(q) = q - 120 $ (in hundreds of dollars), where $ q $ represents the number of batches of biscuits.
- A fixed cost: $ FC = 3620 $ (in dollars).

### 1. **Total Cost Function (TC):**

The total cost is the sum of the fixed and variable costs:

\[
TC(q) = FC + VC(q)
\]

Substitute the given values:

\[
TC(q) = 3620 + 100(q - 120)
\]
\[
TC(q) = 3620 + 100q - 12000
\]
\[
TC(q) = 100q - 8380
\]

So, the total cost function is:

\[
TC(q) = 100q - 8380
\]

---

### 2. **Minimizing the Total Cost:**

To minimize the total cost, we typically need to find the critical points by differentiating the total cost function and setting it to zero. However, since $ TC(q) $ is a linear function, it does not have a minimum or maximum unless constrained. It is not possible to minimize the total cost with a linear function, and the more batches produced, the higher the cost.

Since the problem suggests finding the number of biscuits that minimizes the total cost, this likely refers to the "average cost" function instead. Let's proceed to that next.

---

### 3. **Average Cost Function (AC):**

The average cost function is given by dividing the total cost by the number of batches $ q $:

\[
AC(q) = \frac{TC(q)}{q} = \frac{100q - 8380}{q}
\]

Simplifying:

\[
AC(q) = 100 - \frac{8380}{q}
\]

---

### 4. **Minimizing the Average Cost:**

To minimize the average cost function, we take its derivative with respect to $ q $, set it equal to zero, and solve for $ q $:

\[
AC'(q) = -\frac{8380}{q^2}
\]

Setting $ AC'(q) = 0 $ leads to no solution. However, since $ AC'(q) $ is negative, the average cost decreases as $ q $ increases. The more batches we produce, the lower the average cost per batch.

---

### 5. **Calculating Average Cost for 1 Batch:**

If $ q = 1 $, we substitute into the average cost function:

\[
AC(1) = 100 - \frac{8380}{1} = 100 - 8380 = -8280
\]

This doesn't seem practical for average cost, meaning we might need more clarity on the context, but the calculations proceed logically.

Would you like more details on this process? 

---

### Follow-up questions:

1. How does increasing $ q $ affect the average cost?
2. What happens to the total cost function when we increase $ q $ beyond 100?
3. Can you visualize the average cost function graphically?
4. What other costs (e.g., overhead) might affect this total cost function in reality?
5. How would constraints (e.g., maximum capacity) influence the batch number?

---

**Tip:** When working with cost functions, always consider both total and average cost behavior for better optimization insights.

A biscuit factory's variable cost per month in hundreds of dollars are given by the equation q - 120, where q represents batches of biscuits. The company incurs a fixed cost of $3620. Determine the total cost function. How many biscuits need to be produced to minimize the total cost? Now determine the average cost function. Based on the batches of biscuits that minimizes the total cost, what is the average cost for biscuits? (round to the nearest dollar).

Explore how to determine and minimize the total and average cost functions for a biscuit factory using calculus concepts. This example focuses on optimizing costs with derivative analysis. Suitable for students in Grades 11-12.

Math Problem Statement

A biscuit factory's variable cost per month in hundreds of dollars are given by the equation q

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation