HW #3: Problem #2 A bricklayers union agrees to furnish 50 bricklayers to a shop

Question

HW #3: Problem #2

A bricklayers union agrees to furnish 50 bricklayers to a shopping mall builder. The bricklayers are classified in three categories according to skill: low, medium, and high. The union requires that the total number of medium and high skilled bricklayers are at least four times the number of low skilled. The average number of bricks laid per hour for each skill level is: low, 40 per hour; medium, 60 per hour; high, 75 per hour. The builder knows that the bricklayer crew must lay 3100 bricks per hour to stay on schedule. If the wages per hour are $10, $15, $18 for low, medium, and high skills respectively, how many of each type should be hired to minimize total hourly wages?

Please show all work, even if using a calculator. Go step, by step. Thanks.

Explanation / Answer

let x = the number of low skilled brick layers
let y = the number of medium skilled brick layers
let z = the number of high skilled brick layers

1)x+y+z = 50

2)y + z >= 4x

3) 40x + 60y +75z >=3100

x>=0, y>=0, z>=0 obviously because we can't have a negative amount of workers.

In order to solve this equation we need to simply find the intersection point of all three equations.

Because the intersection of three planes is a point this will be where we minimize the total hourly wages. Moreoever we see that the first equation is not an inequality but rather an equality so therefore the intersection point has to be the solution.

We can place the three equations in a matrix

1    1   1    50

-4   1 1    0

40 60 75 3100

Now we can perform Gauss Jordian elimination:

Add (4 * row1) to row2
1     1     1     50
0     5     5     200
40     60     75     3100

Add (-40 * row1) to row3
1     1     1     50
0     5     5     200
0     20     35     1100

Divide row2 by 5
1     1     1     50
0     1     1     40
0     20     35     1100

Add (-20 * row2) to row3
1     1     1     50
0     1     1     40
0     0     15     300

Divide row3 by 15
1     1     1     50
0     1     1     40
0     0     1     20

Add (-1 * row3) to row2
1     1     1     50
0     1     0     20
0     0     1     20

Add (-1 * row3) to row1
1     1     0     30
0     1     0     20
0     0     1     20

Add (-1 * row2) to row1
1     0     0     10
0     1     0     20
0     0     1     20

Therefore we have 10 low skilled workers, 20 medium skilled workers and 20 high skilled workers

10(10) + 15(20) +18(20)= 760 dollars

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