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Linear Programming

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The Dub-Dub and Dub Company produces and markets three lines of WEB page designs: A, B, and C; A is a standard WEB page design and B and C are professional WEB page designs. The manufacturing process for the WEB page designs is such that two development operations are required - all WEB page designs pass through both operations. Each WEB page design requires 3 hours of development time in Operation 1. In Operation 2 WEB page design A requires 2 hours of development time; WEB page design B requires 4 hours; and WEB page design C requires 5 hours. Operation 1 has 50 hours of development time per week and Operation 2 has sufficient manpower to support 80 hours of development per week. The market group for Dub-Dub and Dub has projected that the demand for the standard WEB page design will be no more than 25 per week. Because WEB page designs B and C are similar in quality, the combined demand for those WEB page designs has been forecast - the total demand is ten or more, but not more than 30 per week. The sale of WEB page design A results in $7 profit while WEB page design B and C provide $8 and $8.5 profits respectively. How many of WEB page designs A, B, and C should be produced weekly if the company seeks to maximize profits? Formulate the problem as a standard LP model.

See attachment
a. Graph the problem.
b. What is the optimal solution?
c. What would the solution be if the third constraint were removed from the problem?

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1. Solution. Assume that we should produce WEB page designs A x pages, B y pages, and C z pages weekly. Then the profit function is
c(x,y,z)=7x+8y+8.5z (1)
We formulate this problem as the following Linear ...

Solution provided by:
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
Recent Feedback
  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
  • "excellent work"
  • "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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