BUS 190 Homework 2
For this week’s homework, you will just be asked to set up the linear program as we did in class, and then to rewrite it in the ‘algebraic form’. Homework 3 will involve graphing the constraints and solving the linear program using that graph, so if you want to get a head start on Homework 3 and are sure of your algebraic setup, you can start the graphs. DO NOT INCLUDE graphs here on HW 2.
Units are per bag 
Standard Leather (square feet) 
Premium Leather (square feet) 
Cutting and Sewing Labor hours 
Finishing Labor Hours 
Profit ($) 
Standard 
7 
1 
3 
1 
25 
Deluxe 
0 
9 
4 
2 
65 
Total available 
210 
180 
120 
40 
The decision variables are S for the number of standard bags and D for the number of deluxe bags.
The objective function is to maximize profit. The formula is max (z) = 25S + 65D.
$25 S of standard bags $65 D of deluxe bags
 * +  * = $25S + $65D
1 standard bag 1 deluxe bag
The constraints are:
7 * S standard leather sq ft 0 * D standard leather sq ft
 +  <= 210
1 standard leather sq ft standard leather sq ft
1 * S premium leather sq ft 9 *D premium leather sq ft
 +  <= 180
1 premium leather sq ft premium leather sq ft
3 * S cslabor hours 4 * D cslabor hours
 +  <= 120
1 cslabor hours cslabor hours
1 * S flabor hours 1 * D flabor hours
 +  <= 140
1 flabor hours flabor hours
7S+0D <= 210
1S + 9D <= 180
3S + 4D <= 120
1S + 1D <= 40
Z = 25S + 65D
s.t.
standard leather constraint: 7S+0D <= 210
premium leather constraint: 1S + 9D <= 180
cutting and sewing labor hours constraint: 3S + 4D <= 120
finished labor hours constraint: 1S + 1D <= 40
And
S >= 0, D >= 0
Cost ($) per share 
Risk (unitless) 
Max shares 
Expected annual return ($1,000s) 

Social media 
$50 
0.5 
1 
5 
Infrastructure 
$30 
0.25 
7 

Total 
$100,000 
700 
1000 
The decision variables are S as the number of units purchased in fund S and I as the number of units purchased in fund I.
The objective function is maximizing expected annual return (z) = 5S + 7I
5S * 1 shares 7I * shares
 +  = 5S + 7I
1 share 1 share
5 * S shares $30 * I shares
 +  <= $100,000
1 share 1 share
1S shares <= 1000 shares
$50S + $30I <= $100000
S <= 1000
.5S + 0.25I <= 700
z = maximized expected annual return
z = 5S + 7I
s.t.
cost constraint: 50S + 30I <= 100000
max shares constraint: S <= 1000
risk restraint: .5S + 0.25I <= 700
S >= 0, I >= 0
Set up the profit function for each type of marker, USING DIMENSIONAL ANALYSIS as you would have to do for a linear program objective function where you wanted to maximize profit. It is probably easiest to figure out profit per marker rather than profit per box, but you may use either unit, as long as you are consistent and use the same unit throughout the entire problem. SHOW ALL YOUR WORK TO GET CREDIT. Warning: do NOT skip using units at any stage. Numbers multiplied together without units in intermediate calculations, but with units attached to them at the end will earn you ZERO points, even if your ‘number’ is correct.
Profit function ($) = (1(.2665625+.05))x + ((10/15)(.37190375+.05))y
Max 1A + 1B
5A + 3B ≤ 15
3A + 5B ≤ 15
A, B ≥ 0
Graph the two constraints and find the overall feasible region for the problem. You should use the original large block graph paper I handed out first day of class and use 2 blocks per unit.
Constraint 1
5A + 3B ≤ 15
0A + 3B ≤ 15 and 5A +0B ≤ 15
B = 5
A = 3
Constraint 2
3A + 5B ≤ 15
0A + 5B ≤ 15 and 3A +0B ≤ 15
A = 5
B = 3
Test point for both (0,0)
0*5 + 0*3 ≤ 15, Correct
0*3 + 0*5 ≤ 15, Correct
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