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# Linear Programming Problem: Maximizing Productivity

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The two products that case chemical makes- cs01 and cs02 yield excessive amounts of three different pollutants: a,b,and c. the state government has ordered the company to install and to employ antipollution devices. the following table provides the current daily emissions in kg/1000liters and the maximum of each pollutant allowed in kg.

pollutant======cs01========cs02=====max allowed

A ============ 25 ========40 ====== 43
B ============ 10 ======== 15 ======20
C ============ 80 ======== 60 ======50

the manager of the production department has approved the installation of two antipollution devices. the emissions from each product can be handled by either device in any proportion. ( the emissions are sent through a device only once, that is, the output of one device cannot be the input to the other or back to itself). The following table shows the percentage of each pollutant from each product that is removed by each device.

============device 1============device2
pollutant cs01 cs02 cs01 cs02
A ===========40 === 40 =========== 30==== 20
B ===========60==== 60 =========== 0 ==== 0
C ===========55==== 55 =========== 65==== 80

For example, if the emission from cs01 is sent through device 1, 40% of pollutant A, 60% of pollutant B, and 55 % of pollutant C are removed. Manufacturing considerations dictate that cs01 and cs02 be produced in the ratio of 2 to 1.
Formulate a LP model to determine a plan that maximizes the total daily production ( amount of cs01 plus amount of CS02) while meeting gov requirements.

##### Solution Summary

Linear programming is used tosolve a pollution related problem. The solution is detailed and well presented.

Solution provided by:
###### Education
• 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!"
• "Thank you"
• "Thank you very much for your valuable time and assistance!"

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