Constrained Optimization using Lagrange multipliers with Commerce applications

In summary: Y3JpcHRpb24gVW5zdWJzY3JpYmU6IEhvbGxpcyEgSSdtIGhhdmUgdGhlIHRlc3QgdG9kYXkgZm9yIHRoaXMgcXVlc3Rpb24sIGFueSBodW1hbiBzaG91bGQgYmUgYXByaWwgYWZ0ZXIgdGhlIG9iamVjdGl2ZSBmdW5jdGlvbiBvdXQgb2YgdGhpcyBxdWVzdGlvbiBiZWNhdXNlIG
  • #1
PandaherO
10
0

Homework Statement



Hello! I'm having some difficulty getting the objective function out of this question, any help/hints would be appreciated >.<

Company A prepares to launch a new brand of tablet computers. Their strategy is to release the first batch with the initial price of p_1 dollars, then later lower the price to p_2 dollars to capture more customers. Demand curve follows q=700-p, where p is any price (dollars) and q is the number of ppl (in units of 1000 ppl) who are willing to buy it at price p dollars.

Cost is $300/each to manufacture each tablet, in the first production, and $200 in the 2nd run, due to factory improvements.

Devise a price strategy for Company A to maximize their profit.

Homework Equations



Note ppl who alrady bought the tablet at the higher price will not buy it agian after the price drop. Ppl who buy during the second run are only those willing ot buy at price p_2 but not at price p_1

Profit=Revenues less cost
R=pq

*predicted profit for p_1=500$, p_2=400$ is $60million

The Attempt at a Solution



So there's two sets tablets being made for p1 and p2 and we want to max the profit, and
profit= R-C
or f(p,q)= pq-q*C
given demand curve q=700-p , cost for 1st run, $300 cost for 2nd run $200

p(700-p)-q(300q_1+200q_2) ??
(p_1+p_2)(700-p) -q(300q_1+200q_2)??

^I think I'm missing something for the objective function :S and I'm not quite sure where the hint (p1,p2)=(500,400) is 60million comes in...
 
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  • #2
PandaherO said:

Homework Statement



Hello! I'm having some difficulty getting the objective function out of this question, any help/hints would be appreciated >.<

Company A prepares to launch a new brand of tablet computers. Their strategy is to release the first batch with the initial price of p_1 dollars, then later lower the price to p_2 dollars to capture more customers. Demand curve follows q=700-p, where p is any price (dollars) and q is the number of ppl (in units of 1000 ppl) who are willing to buy it at price p dollars.

Cost is $300/each to manufacture each tablet, in the first production, and $200 in the 2nd run, due to factory improvements.

Devise a price strategy for Company A to maximize their profit.

Homework Equations



Note ppl who alrady bought the tablet at the higher price will not buy it agian after the price drop. Ppl who buy during the second run are only those willing ot buy at price p_2 but not at price p_1

Profit=Revenues less cost
R=pq

*predicted profit for p_1=500$, p_2=400$ is $60million


The Attempt at a Solution



So there's two sets tablets being made for p1 and p2 and we want to max the profit, and
profit= R-C
or f(p,q)= pq-q*C
given demand curve q=700-p , cost for 1st run, $300 cost for 2nd run $200

p(700-p)-q(300q_1+200q_2) ??
(p_1+p_2)(700-p) -q(300q_1+200q_2)??

^I think I'm missing something for the objective function :S and I'm not quite sure where the hint (p1,p2)=(500,400) is 60million comes in...

The two formulas you wrote above (before the ?? signs) make no sense at all. One of them has a p in it, but there is no p: there are only p_1 and p_2. The other has a q in it (as well as p), but there is no q: there are only q_1 and 1_2. So, instead of writing down random formulas, stop and *think*, and approach the problem systematically. If price p_1 is given, how many tablets are sold (that is q_1)? What it the revenue? What is the cost? After the price is cut to p_2, what is q_2? What is the revenue? What is the cost? Altogether, what is the total revenue and the total cost?

What is the relevance of the statement that people who bought at price p_1 will not then buy again at price p_2?

RGV
 

1. What is constrained optimization?

Constrained optimization is a mathematical process for finding the maximum or minimum value of a function, subject to certain constraints or limitations.

2. What are Lagrange multipliers?

Lagrange multipliers are mathematical tools used in constrained optimization to incorporate constraints into the objective function. They help to find the optimal solution by creating a new function that combines the objective function and the constraints.

3. How are Lagrange multipliers used in commerce applications?

Lagrange multipliers are used in commerce applications to find the optimal solution for a given problem that involves constraints. For example, in inventory management, the objective may be to minimize costs while staying within certain budget constraints. Lagrange multipliers can help find the optimal solution that balances these objectives and constraints.

4. What are some common applications of constrained optimization in commerce?

Some common applications of constrained optimization in commerce include supply chain management, production planning, resource allocation, and portfolio optimization. These problems often involve multiple constraints and objectives, making constrained optimization an essential tool for finding the most efficient and effective solution.

5. What are the benefits of using constrained optimization in commerce?

Constrained optimization allows businesses to find the optimal solution for complex problems with multiple constraints, which can lead to cost savings, increased efficiency, and better decision-making. It also helps to account for real-world limitations and trade-offs, making the solution more practical and realistic.

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