A freight company has shipping orders for two products. The first product has a unit volume of 10 cu ft and weighs 50 lb. The second product's unit volume is 3 cu ft, and it weighs 40 lb. If the company's trucks have 2,300 cu ft of space and can carry 21,000 lb, how many units of each product can be transported in a single shipment with one truck using the entire volume and weight capacity?


To determine how many units of each product can be transported using the entire volume and weight capacity of the truck, we need to set up and solve a system of equations.

### Variables:
- Let \( x \) be the number of units of the first product.
- Let \( y \) be the number of units of the second product.

### Given:
1. The volume of the first product per unit is 10 cubic feet.
2. The weight of the first product per unit is 50 pounds.
3. The volume of the second product per unit is 3 cubic feet.
4. The weight of the second product per unit is 40 pounds.
5. The total volume capacity of the truck is 2,300 cubic feet.
6. The total weight capacity of the truck is 21,000 pounds.

### System of Equations:
1. Volume constraint:
   \[
   10x + 3y = 2300
   \]
2. Weight constraint:
   \[
   50x + 40y = 21000
   \]

We now need to solve this system of linear equations to find the values of \( x \) and \( y \).

### Step 1: Solve the Volume Equation for \( y \)
\[
10x + 3y = 2300
\]
\[
3y = 2300 - 10x
\]
\[
y = \frac{2300 - 10x}{3}
\]

### Step 2: Substitute into the Weight Equation
Substitute \( y = \frac{2300 - 10x}{3} \) into the weight equation:

\[
50x + 40\left(\frac{2300 - 10x}{3}\right) = 21000
\]
\[
50x + \frac{40(2300 - 10x)}{3} = 21000
\]
Multiply through by 3 to eliminate the fraction:

\[
150x + 40(2300 - 10x) = 63000
\]
Distribute the 40:

\[
150x + 92000 - 400x = 63000
\]
Combine like terms:

\[
-250x + 92000 = 63000
\]
\[
-250x = 63000 - 92000
\]
\[
-250x = -29000
\]
\[
x = \frac{-29000}{-250} = 116
\]

### Step 3: Substitute \( x = 116 \) back into the Volume Equation
Now substitute \( x = 116 \) into the volume equation to find \( y \):

\[
10(116) + 3y = 2300
\]
\[
1160 + 3y = 2300
\]
\[
3y = 2300 - 1160
\]
\[
3y = 1140
\]
\[
y = \frac{1140}{3} = 380
\]

### Conclusion:
The company can transport **116 units** of the first product and **380 units** of the second product in a single shipment, using the entire volume and weight capacity of the truck.

Would you like to go over any details, or do you have any questions?

---

Here are 5 additional questions related to this problem:

1. How would the solution change if the truck had a weight capacity of 25,000 lb instead?
2. What if the volume per unit of the second product increased to 5 cu ft? How would that affect the number of units transported?
3. Could the company transport exactly 300 units of the second product? If so, how many units of the first product could be transported?
4. What if the company wanted to minimize the number of units transported while still using the entire truck's capacity?
5. How would introducing a third product with different volume and weight requirements affect the solution?

**Tip:** When solving systems of equations involving constraints like volume and weight, always check if the solution makes practical sense in real-world terms, such as whether it's feasible to have fractional units.

Explore how to determine the number of units of each product a freight company can transport in a single truck shipment. This problem involves solving a system of linear equations considering volume and weight constraints, with detailed step-by-step calculations. Suitable for Grades 7-9 students learning about linear equations and practical applications in logistics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation