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lesantik [10]
2 years ago
14

Candy Kane Cosmetics (CKC) produces Leslie Perfume, which requires chemicals and labor. Two production processes are available:

Process 1 transforms 1 unit of labor and 2 unites of chemicals into 3 oz of perfume. Process 2 transforms 2 units of labor and 3 units of chemicals into 5 oz of perfume. It costs CKC $3 to purchase a unit of labor and $2 to purchase a unit of chemicals. Each year, up to 20,000 units of labor and 35,000 units of chemicals can be purchased. In the absence of advertising, CKC believes it can sell 1,000 oz of perfume. To stimulate demand for Leslie Perfume, CKC can hire the lovely model Jenny Nelson. Jenny is paid $100/hour. Each hour Jenny works for the company is estimated to increase the demand for Leslie Perfume by 200 oz. Each ounce of Leslie Perfume sells for $5. Formulate a linear program to determine how CKC can maximize profits.
Computers and Technology
1 answer:
Vilka [71]2 years ago
3 0

Answer:

  1. Divide the resources into three parts using the corresponding process 1, process 1, and process 2 formats to maximize the use of the resources.
  2. Get the expected revenue by calculating the product of the total perfume in ounce and the price of an ounce of perfume.
  3. Increase the advertisement hours of the product.
  4. subtract the advert fee from the generated revenue to get the actual revenue.
  5. subtract the cost of production from the actual revenue to get the actual profit.

Explanation:

The get maximum profit, all the resources must be exhausted in production. The labor is divided into a ratio of 1:1:2 ( which is 5000, 5000, 1000), while the chemical units are in the ratio of 2:2:3 (10000,10000,15000). This would produce in each individual processes; 15000, 15000 and 25000 oz, which is a total of 55000 oz of perfume.

The expected revenue is $275000. If 1000oz from the 55000oz of perfume is sold without advertisement, model Jenny's awareness of the perfume increases the demand by 200oz per hour, therefore, 24hours would field 4800oz demanded, which would only take 270 hours to distribute all remaining perfumes.

The cost of production would be $130000 for labor and chemical resources plus the advert cost of $27000 ( 270 hours by 100) which is a total cost of $157000. The actual profit is $118000 ( $275000 - $157000).

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Answer:

The benefits of having the operational data ware-house are as follows:

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The challenges of the operational warehouse repository are as follows:

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Under Juran's Law, whenever a problem occurs, what percentage of the time is the problem the result of a system/process error?
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Answer:

C. 85 percent

Explanation:

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4 0
2 years ago
A list named parking_tickets has been defined to be the number of parking tickets given out by the city police each day since th
Murrr4er [49]

Answer:

The needed code is highlighted from the current text. The rest of the code is for the simplicity purpose and to show the functionality of the logic that is used to find the most of the tickets sold till date.

Please do go through it

Explanation:

# import the required packages

import datetime

import random

# define the parking_tickets

parking_tickets = []

# set the 1'st day of the year in d1

d1 = datetime.datetime(2016,1,1)

# set the current day

d2 = datetime.datetime.now()

# calculate number of days till date

totalDays = (d2-d1).days

# fill the number of tickets sold each day randomly

# into the parking_tickets till date

for i in range(totalDays):

    randValue = random.randint(1, 100)

    parking_tickets.append(randValue)

# print the list of tickets sold in each day

print("The list of tickets sold over starting from January 1 to till day are: \n",parking_tickets)

# the required actual code

# define a variable to hold the most of the tickets sold

most_tickets = 0

# logic to find the tickets that sold the most

for k in range(len(parking_tickets)):

    # condition to check most_tickets value is lesser than

    # parking_tickets at index k(each day)

    if most_tickets < parking_tickets[k]:

         # if it is less than the parking_tickets at index k

         # then set the value to the most_tickets

         most_tickets = parking_tickets[k]

# print the most tickets sold

print("Most of the tickets that are sold with in date is ", most_tickets)

5 0
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