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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