Problem 27
Question
Production Planning. \(\quad\) A manufacturer builds racing bikes and mountain bikes, with the per-unit manufacturing costs shown in the table. The company has budgeted \(\$ 26,150\) for materials and \(\$ 31,800\) for labor. How many bicycles of each type can be built? $$ \begin{array}{|l|c|c|} \hline \text { Model } & \text { Cost of materials } & \text { Cost of labor } \\\ \hline \text { Racing } & \$ 110 & \$ 120 \\ \text { Mountain } & \$ 140 & \$ 180 \\ \hline \end{array} $$
Step-by-Step Solution
Verified Answer
The manufacturer can build 85 racing bikes and 120 mountain bikes.
1Step 1: Define Variables
Let \( x \) be the number of racing bikes and \( y \) be the number of mountain bikes. Our goal is to find how many of each type can be manufactured given the budget constraints.
2Step 2: Set Up Equations for Materials
The cost equation for materials is \( 110x + 140y = 26150 \). This represents the budget limit for materials.
3Step 3: Set Up Equations for Labor
The cost equation for labor is \( 120x + 180y = 31800 \). This equation represents the budget limit for labor.
4Step 4: Solve the System of Equations
We use the method of substitution or elimination to solve the system of equations: 1. \( 110x + 140y = 26150 \) 2. \( 120x + 180y = 31800 \). Simplifying 1 gives: \( 11x + 14y = 2615 \) and simplifying 2 gives \( 12x + 18y = 3180 \). Multiply the first equation by 6 and the second by 5: \[ 66x + 84y = 15690 \] \[ 60x + 90y = 15900 \]. Solve using elimination.
5Step 5: Eliminate and Solve for One Variable
Subtract the equations: \( (66x + 84y) - (60x + 90y) = 15690 - 15900 \) which simplifies to: \( 6x - 6y = -210 \). Divide by 6 to get: \( x - y = -35 \).
6Step 6: Substitute to Find Values
From \( x - y = -35 \), express \( x \) as \( x = y - 35 \). Substitute \( x = y - 35 \) into \( 11x + 14y = 2615 \):\( 11(y-35) + 14y = 2615 \).Simplify to get: \( 11y - 385 + 14y = 2615 \), which further simplifies to \( 25y = 3000 \). Solving gives \( y = 120 \).
7Step 7: Solve for the Other Variable
Substitute \( y = 120 \) back into \( x = y - 35 \) to find \( x \). Thus, \( x = 120 - 35 = 85 \).
8Step 8: Verify the Solution
Recalculate both equations with \( x = 85 \) and \( y = 120 \): 1. \( 110(85) + 140(120) = 26150 \) - correct. 2. \( 120(85) + 180(120) = 31800 \) - correct. Both budget constraints are met.
Key Concepts
Production PlanningSystems of EquationsBudget Constraints
Production Planning
Production planning is a key process in manufacturing that involves determining how much of each product should be produced to meet customer demand and business constraints. In our exercise, we are planning the production of racing and mountain bikes. The manufacturer needs to consider several factors:
This also involves creating a strategy that considers both product costs and available production capabilities.
- The cost of materials for each bike model
- The cost of labor to produce each bike model
- The total budget available for materials and labor
This also involves creating a strategy that considers both product costs and available production capabilities.
Systems of Equations
Systems of equations are a set of two or more equations with the same variables. In our context, these equations represent the relationships between the number of bikes produced and the cost constraints for materials and labor.
In the example, our system of equations starts with:
In the example, our system of equations starts with:
- Material costs: \(110x + 140y = 26150\)
- Labor costs: \(120x + 180y = 31800\)
Budget Constraints
Budget constraints are limits on the spending capacity of a business. They define how much can be spent in particular areas, such as materials and labor in our exercise. Meeting budget constraints is crucial to sustainable business operations.
In the bike production exercise, we tackled the following budget constraints:
In the bike production exercise, we tackled the following budget constraints:
- Materials cost cannot exceed \(\\(26,150\)
- Labor cost cannot exceed \(\\)31,800\)
Other exercises in this chapter
Problem 26
Use matrices to solve each system of equations. $$ \left\\{\begin{array}{l} x+y=3 \\ x-y=-1 \end{array}\right. $$
View solution Problem 26
Solve each system. $$ \left\\{\begin{array}{l} 3 y+z=-1 \\ -x+2 z=-9+6 y \\ 9 y+3 z=-9+2 x \end{array}\right. $$
View solution Problem 27
Evaluate each determinant. $$ \left|\begin{array}{lll} 3 & 2 & 1 \\ 4 & 1 & 2 \\ 5 & 3 & 1 \end{array}\right| $$
View solution Problem 27
Use matrices to solve each system of equations. $$ \left\\{\begin{array}{l} 2 x+y=1 \\ x+2 y=-4 \end{array}\right. $$
View solution