Myisha coleman week 6 cyu b what is the total market

Info icon This preview shows pages 7–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Myisha Coleman Week 6 CYU b) What is the total market demand for polyglue at the price established by Alchem in Part (a)? How much of total demand do the follower firms supply? Part (a) Alchem's (L) profit-maximizing output occurs where: MR L = MC L MR L is found as follows: MR L = d(TR L )/dQ L TR L is given by the following expression: TR L = P  Q L Also, Q L is given by: Q L = Q T Q F Using the total demand function, one can solve for Q T : P = 20,000 4Q T or Q T = 5000 .25P In order to find Q F , one notes that Alchem lets the follower firms (F) sell as much polyglue as they wish at the given market price (P). Therefore, the follower firms are faced with a horizontal demand function and hence: MR F = P In order to maximize profits, the follower firms will operate where: MR F = MC F , or where: P = 2000 + 4Q F Solving for Q F yields: Q F = .25 P 500 Substituting Q F and Q T into the expression above for Q L gives: Q L = (5000 .25P) (.25P 500) = 5500 .50P Solving for P gives: P = 11,000 2Q L
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Myisha Coleman Week 6 CYU Substituting P into the TR L expression above gives: TR L = (11,000 2Q L )Q L = 11,000Q L 2Q L 2 MR L is therefore equal to: MR L = 11,000 4Q L Setting MR L = MC L gives: 11,000 4Q L = 5000 + 5Q L so Q L * = 666.7. Substituting this value into the expression for P gives: P* = 11,000 2(666.7) = $9,666.70. Part (b) From the expression above for Q T , one obtains: Q T * = 5000 .25(9666.7) = 2,583.3. The followers supply the rest: Q F * = .25(9666.7) 500 = 1,916.7.
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern