How Does Car Type and Investment Limit Affect Dealership Profit Maximization?

In summary, the manager of an automobile dealership must decide how many cars to order for the new model year in order to maximize his profit. There are two types of cars- midsize cars and compact cars. The selling price and costs are listed in the table below. The manager wants at least 10 of each type, but no more than 50 of the midsize cars and no more than 60 of the compact cars. The manager assumes that as long as he meets these quantity limits, he will be able to sell all the cars that he orders. A linear programming model for this problem can be written out algebraically, and a spreadsheet model can be solved using Solver. The selling price of the midsize car might be as low as
  • #1
jianfeyyu
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Homework Statement

A manager of an automobile dealership must decide how many cars to order for the new model year in order to maximize his profit. There are two types of cars: midsize cars and compact cars. The selling price and costs are listed in the following table:

Car type -- Selling price -- Cost

Midsize -- 25,000−−18,178

Compact -- 18,000−−12,000
No more than $790,000 can be invested. The manager wants at least 10 of each type, but no more than 50 of the midsize cars and no more than 60 of the compact cars. (The manager assumes that as long as he meets these quantity limits, he will be able to sell all the cars that he orders.)

1.Formulate a linear programming model for this problem. That is, write out an algebraic statement of the model.
2.Create a spreadsheet model for this problem and solve it using Solver. What is the optimal solution and resulting profit?
3.The selling price of the midsize car might be as low as 22,000 or as high as 27,000. How does your solution found in #2 change with these prices?

Homework Equations

max: z = (25000 - 18178)x + (18000 - 12000)y
= 6822x+6000y
18178x + 12000y <= 790000
x>=10
x<=50
y>=10
Y<=60
x,y>=0

The Attempt at a Solution

I tired to set up the equation already. However, I spent a whole day to figure out the Constraints' part in the Excel. Can someone help?
 

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  • #2
Please fix your picture
 
  • #3
what's wrong with the picture?
 
  • #4
jianfeyyu said:
what's wrong with the picture?
It's not shown.
Use the attachment function to attach the picture.It seems the site(MathHelpForum) does not allow non-members to see the picture!
 
  • #5
Do you still need help? Based on the edit, it appears that you have correctly solved the problem.
 
  • #6
yes, i think the D 22- D 25 should not be zero. Can you show me your calculation for B22 to D 25 and the excel solever?
 
  • #7
jianfeyyu said:
yes, i think the D 22- D 25 should not be zero.
Two questions with regard to this:
1. What do you mean by this?
2. Why do you think this is the case?

Can you show me your calculation for B22 to D 25 and the excel solever?
No, that would be against the rules of this site.

What happens in your spreadsheet when you change the selling price of the midsize car from $25,000 to $22,000? When you change it from $25,000 to $27,000?

Here are two more questions that might help you understand what's going on:
- What happens when you change the midsize selling price from $25,000 to $28,000?
- What happens when you change it from $25,000 to $27,267?

If you've set up your spreadsheet nicely, all you have to do is change one value and rerun the solver to find the answers to each of these four questions.

It turns out that this problem is easy to solve by a simple calculator. A number of the constraints are superfluous. Since you already know that your x and y must each be at least ten means the constraints that x and y are non-negative are superfluous. If the manager orders 60 compacts, that leaves room for only 3.85 midsize cars in the $790,000 budget. If the manager orders 50 midsize cars, he's already blown the $790,000 budget by $118,900. The 60 car limit on compacts and the 50 car limit on midsize cars are both superfluous. That leaves three constraints: The $790,000 budget and the desires to have at least ten midsize cars and and least ten compacts.

This means the feasibility region is a triangle with one of the vertices representing ordering the minimum, and the other two vertices represent ordering ten cars of one type and maxing out that $790,000 budget on the other type. That vertex with ten cars each obviously is not the solution.

Note that changing the selling price doesn't change the feasibility region one iota. The same feasibility region applies to all of the subquestions.

Note also that your solution has the manager buying 50.685 compacts. How do you buy 50 cars plus 68.5% of another car?
 

FAQ: How Does Car Type and Investment Limit Affect Dealership Profit Maximization?

What is linear programming?

Linear programming is a mathematical method used to optimize a linear objective function, subject to a set of linear constraints. It is often used to maximize profits, minimize costs, or find the most efficient use of resources.

How do I set up a linear programming problem?

To set up a linear programming problem, you will need to define the objective function, which is the quantity you want to optimize, and the constraints, which are the limitations on the variables in the problem. You can then use mathematical techniques to solve for the optimal solution.

What are the key assumptions of linear programming?

The key assumptions of linear programming include linearity, proportionality, additivity, divisibility, and certainty. These assumptions allow for the use of mathematical techniques to solve the problem and find an optimal solution.

What are the benefits of using linear programming?

Linear programming allows for the optimization of a problem with multiple variables and constraints. It can help individuals and organizations make more informed decisions, maximize profits, minimize costs, and improve efficiency.

What are some common applications of linear programming?

Linear programming has many applications in various industries, including transportation, finance, manufacturing, and supply chain management. It can be used to solve problems such as production planning, inventory management, resource allocation, and scheduling.

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