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tresset_1 [31]
3 years ago
12

Each machine must be run by one of 19 cross-trained workers who are each available 35 hours per week. The plant has 10 type 1 ma

chines available, 6 type 2 machines available, and 8 type 3 machines available. Products 1, 2, and 3 contribute $90, $120, and $150, respectively, in marginal profit per unit produced.

Business
1 answer:
Mrac [35]3 years ago
4 0

Answer:

The Linear programming model is given as below

Profit Function: P=90X+120Y+150Z

Constraints:

2X+2Y+Z\leq 400

3X+4Y+6Z\leq 240

4X+6Y+5Z\leq 320

\dfrac{2X+2Y+Z}{40}\leq 10

\dfrac{3X+4Y+6Z}{40}\leq 6

\dfrac{4X+6Y+5Z}{40}\leq 8

\dfrac{2X+2Y+Z}{35}+\dfrac{3X+4Y+6Z}{35}+\dfrac{4X+6Y+5Z}{35}\leq 19

Explanation:

As the question is not complete, the complete question is found online and is attached herewith.

Let the number of product 1 to be produced is X, that of product 2 is Y and product 3 is Z

so  the maximizing function is the profit function which is given as

P=90X+120Y+150Z

Now as the number of hours in a week are 40 and there are a total of 10 type 1 machines so the total number of machine 1 hours are 40*10=400 hours

As from the given table product 1 uses 2 machine hours of machine 1, product 2 uses 2 machine hours of machine 1 and product 3 uses 1 hour of machine 1 so

2X+2Y+Z\leq 400

Now as the number of hours in a week are 40 and there are a total of 6 type 2 machines so the total number of machine 2 hours are 40*6=240 hours

As from the given table product 1 uses 3 machine hours of machine 2, product 2 uses 4 machine hours of machine 2 and product 3 uses 6 hour of machine 2 so

3X+4Y+6Z\leq 240

Now as the number of hours in a week are 40 and there are a total of 8 type 3 machines so the total number of machine 3 hours are 40*8=320 hours

As from the given table product 1 uses 4 machine hours of machine 3, product 2 uses 6 machine hours of machine 3 and product 3 uses 5 hour of machine 3 so

4X+6Y+5Z\leq 320

Now as the machine 1 is used as 2X+2Y+Z in a week and the week is of 40 hours so the number of machines to be used are given as

\dfrac{2X+2Y+Z}{40}\leq 10

Now as the machine 2 is used as 3X+4Y+6Z in a week and the week is of 40 hours so the number of machines to be used are given as

\dfrac{3X+4Y+6Z}{40}\leq 6

Now as the machine 3 is used as 4X+6Y+5Z in a week and the week is of 40 hours so the number of machines to be used are given as

\dfrac{4X+6Y+5Z}{40}\leq 8

Now the workers are available for 35 hours so the worker available at the machine 1 is given as

\dfrac{2X+2Y+Z}{35}

That of machine 2 is given as

\dfrac{3X+4Y+6Z}{35}

That of machine 3 is given as

\dfrac{4X+6Y+5Z}{35}

As the total number of workers is 19 so the constraint is given as

\dfrac{2X+2Y+Z}{35}+\dfrac{3X+4Y+6Z}{35}+\dfrac{4X+6Y+5Z}{35}\leq 19

So the Linear programming model is given as below

Profit Function: P=90X+120Y+150Z

Constraints:

2X+2Y+Z\leq 400

3X+4Y+6Z\leq 240

4X+6Y+5Z\leq 320

\dfrac{2X+2Y+Z}{40}\leq 10

\dfrac{3X+4Y+6Z}{40}\leq 6

\dfrac{4X+6Y+5Z}{40}\leq 8

\dfrac{2X+2Y+Z}{35}+\dfrac{3X+4Y+6Z}{35}+\dfrac{4X+6Y+5Z}{35}\leq 19

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