At Long John Silver’s (LJS), Platter 1 comes with 2 pieces of fish, 3 pieces of chicken, and 5 fried shrimp, while Platter 2 comes with 3 pieces of fish, 1 piece of chicken, and 8 fried shrimp. Platter 1 sells for 8 dollars and platter 2 sells for 10 dollars. If LJS has access to 1000 pieces of fish, 1200 pieces of chicken, and 2600 fried shrimp, what is the optimal number of platters LJS should make in order to maximize its profit? Note that the constraints relate to (please use these in order) fish, chicken, and shrimp.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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  1. At Long John Silver’s (LJS), Platter 1 comes with 2 pieces of fish, 3 pieces of chicken, and 5 fried shrimp, while

    Platter 2 comes with 3 pieces of fish, 1 piece of chicken, and 8 fried shrimp. Platter 1 sells for 8 dollars and platter 2 sells for 10 dollars. If LJS has access to 1000 pieces of fish, 1200 pieces of chicken, and 2600 fried shrimp, what is the optimal number of platters LJS should make in order to maximize its profit? Note that the constraints relate to (please use these in order) fish, chicken, and shrimp.

  2. Note that problem 7 had three constraints (excluding the non-negativity constraints), and setting two pairs of those constraints to equality led to two extreme points on the feasible region. What INFEASIBLE point was at the intersection of the other pair of constraints?

  3. In the optimal solution to problem 7, the optimum was found at the intersection of two of the constraints that were not non-negativity constraints. Call these “binding” constraints. One of those two constraints represented the number of available pieces of fish. What would happen if you increased the availability of the number of pieces of fish by 10? In other words, solve a nearly identical linear program, but increase the right- hand side of one of your inequalities to 1010. What is the new optimal value?

  4. Now increase that inequality to 1100 and see what happens. What is the optimal value now?

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