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

Question

A bicycle manufacturer builds one-, three- and ten-speed models. The bicycles need both aluminum and steel. The company has available 48,420 units of steel and 36,645 units of aluminum. The one-, three, and ten-speed models need, respectively, 12, 16 and 20 units of steel and 12, 9, and 15 units of aluminum. How many of each type of bicycle should be made in order to maximize profit if the company makes $6 per one-speed bike, $8 per three-speed, and $20 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

Let manufacturer builds x , y, z bicycles of One, Three and Ten models respectively.

We can formulate the given problem as following

Max
6x + 8y + 20z
Subject to
12x + 16y + 20z <= 48420
12x + 9y + 15z <= 36645

x, y, z >=0

Simplex method

Using the slackes variables

Max
6x + 8y + 20z
Subject to
12x + 16y + 20z + S1 = 48420
12x + 9y + 15z + S2 = 36645

x, y, z, S1, S2 >=0

Cj - Zj is maximum postive for 3rd column i.e. 20, So entering variable is z.

Minimum ratio is for first row, so leaving variable is S1.

=> Pivot is 20

Row opertaions will be

R1(new) = R1(old) /20

R2(new) = R2(old) - 15 R1(new)  

all Cj-Zj <= 0

=> Optimal solution is

x = 0, y = 0, z = 2421

Maximum profit = 6(0) + 8(0) + 20(2421) = 48420

Number of One-speed bicycle = 0

Number of Three-speed bicycle = 0

Number of Ten-speed bicycle = 2421

Table 1 Cj -> 6 8 20 0 0 B CB x y z S1 S2 XB Min Ratio S1 0 12 16 20 1 0 48420 48420/20 =2421 S2 0 12 5 15 0 1 36645 36645/15=2443 Z = 0 Zj 0 0 0 0 0 Cj-Zj 6 8 20 0 0

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