II Content I AssessmentsI CommunicationI Re CH 11 Quiz- Quiz e Left: 0:42:59 Zac

Question

II Content I AssessmentsI CommunicationI Re CH 11 Quiz- Quiz e Left: 0:42:59 Zachary Bradshaw: Attempt 1 Consider the following linear programming problem: MAX 50X60Y s.t. 8x 10Y s 800 1X 1Y S 120 X, Y 20 What is the value of the objective function at the optimal solution? (Hint: solve graphically by hand or formulate in Excel and solve using Solver) 1600 4600 5000 4900 Save Question 8 (1 point of plaasic parts used for A plastic pants supplier produces nutes of machine time. Tvpe Type 1 requires 30 cf machine time 900 mach

Explanation / Answer

Here, we are given the constraints as: ( in equality form )

Multiplying the second equation by 8 and subtracting from the first, we get:

2Y = 800 - 120*8 = -160

Therefore, we get: Y = -80

X = 200

Therefore (200, -80) is the first critical point here. but this cannot be true as both X and Y should be greater than 0.

Next we solve:

These 2 equqations never intersect as they are parallel to each other.

Next, we see:

Multiplying the second equation by 4 and subtracting from the first, we get:

Y = 500 - 480 = 20

Therefore (100,20) is the critical point here. Now checking if it satisfies the other constraint as well.

8X + 10Y = 800 + 200 = 1000 which is not less than 800, therefore this constraint is not satisfied here.

Next we look at the points with either X = 0 or Y = 0. We get the critical points as:

8X + 10Y = 800

At X = 0, Y = 80, obj function value = 60*80 = 4800
At Y = 0, X = 100, obj function value = 50*100 = 5000

This also satisfies all the other constraints. The point (100,0)

X + Y = 100 < 120

4X + 5Y = 400 < 500

Therefore (100,0) is the point of maximization.

The required max value of the objective function here is 5000

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