Salexia Timmerman
QAS 19
Assignment 1
1) d = 800 – 10p
a) When p = $20, d = 800 – 10(20) = 800 – 200 = 600
When p = $70, d = 800 – 10(70) = 800 – 700 = 100
Thus, the firm can sell 600 units at the $20 price per unit and 100 units at the
$70 price per unit.
b) The mathematical model for the total revenue is
TR = dp = (800 – 10p)p
That is, TR = 800p – 10p2
c) When p = $30, TR = 800(30) – 10(30^2) = 24000 – 9000 = 15,000
When p = $40, TR = 800(40) – 10(40^2) = 32000 – 16000 = 16,000
When p = $50, TR = 800(50) – 10(50^2) = 40000 – 25000 = 15,000
Thus, total revenue is maximized at the $40 price.
d) The expected annual demand corresponding to the recommended price is
d = 800 – 10(40) = 800 – 400 = 400
and TR = 800(40) – 10(40^2) = 32000 – 16000 = 16,000
Thus, the expected annual demand is 400 units and the total revenue is $16,000
corresponding to the recommended price.
3) a) The decision variables are
X = Number of units purchased in the stock fund
Y = Number of units purchased in the money market fund
b) Since each unit invested in the stock fund has a risk index of 8, and each unit
invested in the money market fund has a risk index of 3; the total risk index is
8 X + 3Y.
Thus, an objective function that will minimize the total risk index for the
portfolio is
Minimize Z = 8 X + 3 Y
c) Since each unit of stock fund costs $50 and each unit of money market fund
costs $100, the constraint corresponding to the available funds is
50 X + 100 Y ≤ 1,200,000
Since the annual income from stock fund is ($50)(10%) = $5 and the annual
income from the money market fund is ($100)(4%) = $4, the constraint
corresponding to the annual income is
5 X + 4 Y ≥ 60,000
Since at least $300,000 to be invested in the money market, at least
($300,000/$100) = 3,000 units must be invested in the money market. Thus, the
constraint corresponding to the minimum units in money market is
Y ≥ 3,000
Finally, the non-negativity constraints, X, Y ≥ 0.
Thus, the constraints of the problem are
50 X + 100 Y ≤ 1,200,000
5 X + 4 Y ≥ 60,000
Y ≥ 3,000
X, Y ≥ 0
4) The decision variables are
X = Number of gallons of regular gasoline produced
Y = Number of gallons of premium gasoline produced
Since the profit contributions are $0.30 per gallon for regular gasoline and $0.50
per gallon for premium gasoline; the total profit contribution is 0.30 X + 0.50 Y.
Thus, the objective function is
Maximize Z = 0.30 X + 0.50 Y
Since each gallon of regular gasoline contains 0.3 gallons of grade A crude oil
and each gallon of premium gasoline contains 0.6 gallons of grade A crude oil,
the constraint corresponding to the available Grade A crude oil is
0.30 X + 0.60 Y ≤ 18,000
The constraint corresponding to the production capacity is
X + Y ≤ 50,000
The constraint corresponding to the demand for the premium gasoline is
Y ≤ 20,000
Finally, the non-negativity constraints, X, Y ≥ 0.
Thus, the linear programming model is
Maximize Z = 0.30 X + 0.50 Y
Subject to
0.30 X + 0.60 Y ≤ 18,000
X + Y ≤ 50,000
Y ≤ 20,000
X, Y ≥ 0
5) The decision variables are
X = Number of necklaces sold
Y = Number of earrings sold
Since the necklaces make the designer a profit of $25 each piece and the earrings
make a profit of $10 each piece; the designer’s weekly profit is 25 X + 10 Y.
Thus, the objective function is
Maximize Z = 25 X + 10 Y
Since the boutique will buy between 10 and 40 necklaces each week, the
constraint corresponding to requirement of necklaces is
10 ≤ X ≤ 40
Since the boutique bought at least twice the number of earrings as necklaces, the
corresponding constraint is
2Y ≥ X
or
2Y – X ≥ 0
Since the designer cannot make more than 50 pieces of jewelry per week, the
constraint corresponding to this limitation is
X + Y ≤ 50
Finally, the non-negativity constraints, X, Y ≥ 0.
Thus, the linear programming model is
Maximize Z = 25 X + 10 Y
Subject to
10 ≤ X ≤ 40
2Y – X ≥ 0
X + Y ≤ 50
X, Y ≥ 0