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Misha Larkins [42]
3 years ago
9

State the linear programming problem in mathematical terms, identifying the objective function and the constraints. A firm makes

products A and B. Product A takes 2 hours each on machine M; product B takes 4 hours on L and 3 hours on M. Machine L can be used for 8 hours and M for 6 hours. Profit on product A is $9 and $7 on B. Maximize profit.

Mathematics
1 answer:
Sedbober [7]3 years ago
6 0

Answer:

Maximum profit at (3,0) is $27.

Step-by-step explanation:

Let  quantity of  products A=x

Quantity  of products B=y

Product A takes time on machine L=2 hours

Product A takes time on machine M=2 hours

Product B takes time on machineL= 4 hours

Product B takes time on machine M=3 hours

Machine L can used total time= 8hours

Machine M can used total time= 6hours

Profit on product A= $9

Profit on product B=$7

According to question

Objective function Z=9x+7y

Constraints:

2x+3y\leq 6

2x+4y\leq 8

Where x\geq 0, y\geq 0

I equation 2x+3y\leq 6

I equation in inequality change into equality we get

2x+3y=6

Put x=0 then we get

y=2

If we put y=0 then we get

x= 3

Therefore , we get two points A (0,2) and B (3,0) and plot the graph for equation I

Now put x=0 and y=0 in I equation in inequality

Then we get 0\leq 6

Hence, this equation is true then shaded regoin is  below the line .

Similarly , for II equation

First change inequality equation into equality equation

we get 2x+4y=8

Put x= 0 then we get

y=2

Put y=0 Then we get

x=4

Therefore, we get two points C(0,2)a nd D(4,0) and plot the graph for equation II

Point  A and C are same

Put x=0 and y=0 in the in inequality equation II then we get

0\leq 8

Hence, this equation is true .Therefore, the shaded region is below the line.

By graph we can see both line intersect at the points A(0,2)

The feasible region is AOBA and bounded.

To find the value of objective function on points

A (0,2), O(0,0) and B(3,0)

Put A(0,2)

Z= 9\times 0+7\times 2=14

At point O(0,0)

Z=0

At point B(3,0)

Z=9\times3+7\times0=27

Hence maximum value of z= 27 at point B(3,0)

Therefore, the maximum profit is $27.

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