Exercise 1. A company finds that if it produces and sells q units of a product, its total sales revenue in Riyals is 100 pq. If the variable cost per unit is 2 SR and the fixed cost is 1,200 SR, then the values of q for which the profit is zero are:

(a) q = 400 or q = 900

(b)q = 100 or q = 3600

(c) q = 200 or q = 1800

(d)q = 300 or q = 1200

(e) q = 600 or q = 600

Exercise 2. A company will manufacture a total of 6,000 units of product A and product B at two separate locations. Product A has the unit cost of 3 SR and fixed costs of 6,000 SR, while the unit cost for product B is 3.50 SR with 3,000 SR of fixed costs. The minimum number of units of product A that must be produced, such that the total cost does not exceed 28,000 is:

(a) 4000

(b)4545

(c) 4500

(d)3871

(e) 4546

Exercise 3. The average cost for a product has risen by 59.82 SR per year for the years 1990 through 2000. Assume the average cost in the year 1996 was 1128.50 SR. Then, the average cost in the year 2000 was:

(a) 1367.78

(b)1307.96

(c) 1726.70

(d)1407.69

(e) 1476.87

Exercise 4. A new television depreciates 120 SR per year, and it is worth 340 SR after four years. If t is the age of the television in years, a function that describes the value of this television is:

(a) f(t) = -120t+ 820

(b) f(t) = 120t- 820

(c) f(t) = t2 + 5t+ 304

(d) f(t) = t2 + 5t- 376

(e) f(t) = 120t- 140

Exercise 5. The demand function for a product is p = 600 – 3q, where p is the price per unit. The maximum revenue is:

(a) 30,000 SR

(b)23,500 SR

(c) 40,000 SR

(d)55,000 SR

(e) 60,000 SR

Exercise 6. A factory produces fabric made from different fibers. From Cotton, Polyester, and Nylon, the owners want to produce a fabric blend that will cost 3.25 SR per pound to make.

The cost per pound of these three fibers is 4 SR, 3 SR, and 2 SR, respectively. Assume the amount of Nylon is to be the same as the amount of Polyester. Then, the amount of Cotton (per pound) in the final fabric is equal to:

(a) 0.50

(b)0.10

(c) 0.20

(d)0.30

(e) 0.40

Exercise 7. The nonlinear system ( xy == y 2+px1+2

(a) has one solution only

(b)has no solution

(c) has two solutions only

(d)has three solutions only

(e) has infinitely many solutions

Exercise 8. The supply equation for a product is p = 80 q and the demand equation is p = q 20 -3, where p is the price (in Saudi Riyals) per unit. The equilibrium price is

(a) 1

(b)0.5

(c) 1.5

(d)2

(e) 3.5

Exercise 9. The system

8>>><>>>:

x + 2y + 4z = 6

y + 2z = 3

x + y + 2z = 1

(a) has no solution

(b)has infinitely many solutions

(c) has a unique solution

(d)has (0;1;1) as a solution

(e) has three solutions only

Exercise 10. A corporation produces two models of notebook computers: the Bing and the Lambert. Let x be the number of Bing models and y the number of Lambert models produced per week. If the factory can produce at most 750 Bing and Lambert models combined in a week, the inequalities that describe this situation are:

(a) ( xx≥+0 ; y ≤y750 ≥ 0

(b)( xx≥+0 ; y ≥y750 ≥ 0

(c) ( xx++yy≤≥750 0

(d)( xx++yy≥≥750 0

(e) ( xx≥-0 ; y ≤y750 ≥ 0

Exercise 11. A diet is to contain at most 16 units of carbohydrates and at least 20 units of protein. Food A contains 2 units of carbohydrates and 4 units of protein; food B contains 2 units of carbohydrates and 1 unit of protein. Food A costs 1.20 SR per unit and food B costs 0.80 SR per unit. Let x = Number of units of food A and y = Number of units of food B. The linear programming problem to minimize cost Z is:

(a) Minimize Z = 1:2x + 0:8y subject to x + y ≤ 8 ; 4x + y ≥ 20

(b)Minimize Z = 1:2x + 0:8y subject to 2x + 2y ≤ 16 ; 4x + y ≤ 20

(c) Minimize Z = 16x + 20y subject to 2x + 2y ≤ 1:20 ; 4x + y ≤ 0:8

(d)Minimize Z = 1:2x + 0:8y subject to 2x + 4y ≤ 16 ; 2x + y ≥ 20

(e) Minimize Z = 0:8x + 1:2y subject to 2x + 2y ≤ 16 ; 4x + y ≥ 20

Exercise 12. We use the geometric (graphical) approach to optimize Z = 2x + 2y subject to 8>>>>><>>>>>:

x – 2y ≤ 4

2x – y ≥ -4

x + y = 6

x ≥ 0; y ≥ 0

. Then:

(a) Z has a minimum value at (2;4) and (4;2)

(b)Z has no minimum value

(c) Z has no maximum value

(d)Z has a minimum value at (0;4) and (4;0)

(e) Z has a maximum value only at (3;3)

Exercise 13. The initial simplex tableau for a standard maximization problem is set as x y s1 s2 Z

0BBBBBB@

2 3 1 0 0 6

3 1 0 1 0 1

-7 -12 0 0 1 0

1CCCCCCA

: Then, the maximum value of Z is

(a) 12

(b)0

(c) 6

(d)18

(e) 24

Exercise 14. We use the dual and simplex method to solve the following problem:

Minimize Z = 2×1 + 3×2 + 4×3 subject to

8>>><>>>:

x1 + 2×2 + 3×3 ≥ 2

2×1 + 3×2 + 4×3 ≥ 3

x1 + x2 + x3 ≥ 4

Then, Z has a minimum value at (x1;x2;x3) =

(a) (4;0;0)

(b)(0;0;2)

(c) (0;4;0)

(d)(2;0;2)

(e) (0;1;2)