A bicycle manufacturer builds one-, three- and ten-speed models. The bicycle nee

Question

A bicycle manufacturer builds one-, three- and ten-speed models. The bicycle need both aluminum and steel. The company has available 46, 380 units of steel and 59, 475 units of aluminum. The one-, three-, and ten-speed models need, respectively, 12, 16 and 20 units of steel and 20, 15, and 25 units of aluminum. How many of each type of bicycle should be made in order to maximize profit if the company makes $3 per one-speed bike, $4 per three-speed, and $10 per ten-speed. What is the maximum possible profit? Type the number of each type of bicycle that should be made One-speed Three-speed Ten-speed

Explanation / Answer

Solution :

Let x bikes of the one-speed type, y of the three-speed type and z of the ten-speed type be made to maximize the profit. Then the Linear Programming Problem is:

Maximize P = 3x + 4y + 10z

Subject to

12x + 16y + 20z <= 46380 [Steel availability constraint]

20x + 15y + 25z <= 59475 [Aluminium availability constraint]

x, y, z 0 [Non-negativity constraints]

We'll build the first tableau of the Simplex method.

The leaving variable is P4 and entering variable is P3.

Pivot row (Row 1):
46380 / 20 = 2319
12 / 20 = 0.6
16 / 20 = 0.8
20 / 20 = 1
1 / 20 = 0.05
0 / 20 = 0

Row 2:
59475 - (25 * 2319) = 1500
20 - (25 * 0.6) = 5
15 - (25 * 0.8) = -5
25 - (25 * 1) = 0
0 - (25 * 0.05) = -1.25
1 - (25 * 0) = 1

Row Z:
0 - (-10 * 2319) = 23190
-3 - (-10 * 0.6) = 3
-4 - (-10 * 0.8) = 4
-10 - (-10 * 1) = 0
0 - (-10 * 0.05) = 0.5
0 - (-10 * 0) = 0

The optimal solution value is Z = 23190
X1 = 0
X2 = 0
X3 = 2319

One Speed = 0

Three Speed = 0

Ten Speed = 2319

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