King Fahd University of Petroleum and Minerals
EM 520
H.W.#
By:
Abdulmohsen Al-Dossary
ID #: 200773730
Submission Date: 10/02/2014
Problem 9.1:
Let: X1= number of model A tub to be produced.
X2= number of model B tub to be produced.
Objective function: Maximize profit = $85X1+$70X2
1st constraint: 130 X1+100 X2 ≤ 25,000
2nd constraint: 20 X1+30 X2 ≤ 6,500
3rd and 4th constraint: X1 ≥ 0, X2 ≥ 0
Problem 9.2:
Let: X1= number of benches to be produced.
X2= number of picnic tables to be produced.
Objective function: Maximize profit = $9X1+$20X2
1st constraint: 4X1+7 X2 ≤ 1,200
2nd constraint: 10X1+35 X2 ≤ 5,000
3rd and 4th constraint: X1 ≥ 0, X2 ≥ 0
Problem 9.3:
Let: X1= number of Alpha 6 minicomputers to be produced.
X2= number of Beta 8 minicomputers to be produced.
Objective function: Maximize profit = $1,200X1+$1,800X2
1st constraint: 18X1+25 X2 ≤ 800 (since 5 men working and each is 160 hr)
2nd constraint: X1 ≥ 10
3rd constraint: X2 ≤ 20
Problem 9.4:
1st constraint: 2X1+1X2 ≤ 40
2nd constraint: 1X1+3X2 ≤ 30
3rd and 4th constraint: X1 ≥ 0, X2 ≥ 0
Converting the constrains from inequalities to equalities ( i.e. equations);
2X1+1X2 = 40 and 1X1+3X2 =30
Using excel to graph these constraints, the shaded area is the feasible area;
Problem 9.5:
1st constraint: 5X1+2X2 ≤ 250
2nd constraint: 1X1+2X2 ≤ 150
3rd constraint X1
≤ 40
3rd and 4th constraint: X1 ≥ 0, X2 ≥ 0
Converting the constrains from inequalities to equalities ( i.e. equations);
5X1+2X2 = 250 , 1X1+2X2 =150 and X1 = 40
Using excel to graph these constraints, the shaded area is the feasible area;
Problem 9.6:
The objective function is: Maximize profit = $9X1+$7X2
From the above graph which is taken from problem 9-4;
We observe the shaded feasible region got four corners points which is
highlighted be letters.
Point a: (X1= 0, X2= 10)
Point b: (X1= 0, X2= 0)
Point c: (X1= 20, X2= 0)
Point d: can be calculated by adding the two constrains’ equations as
follows:
2X1+1X2 = 40
-2X1-6X2 = -60 ( multiplied by -2)
0 – 5X2 = -20, so X2 = 4
Substituting in the first equation to get X1 value;
2X1+4=40,
2X1= 36, X1= 18
Point d: (X1= 18, X2= 4)
Finding the profit:
Point a: 9(0)+7(10)= $70
Point b: 9(0)+7(0)= $0
Point c: 9(20)+7(0)= $180
Point d: 9(18)+7(4)= $190
So point d is the answer since this corner pint gives the maximum profit and
X1 should be chosen to be 18 and X2 should be 4 to maximize the profit.
Problem 9.7:
The objective function is: Maximize profit = $2X1+$3X2
From the above graph which is taken from problem 9-4;
We observe the shaded feasible region got five corners points which is
highlighted be letters.
Point a: (X1= 0, X2= 75)
Point b: (X1= 0, X2= 0)
Point c: (X1= 40, X2= 0)
Point d: Can be calculated by solving the two constrains’ equations as
follows:
5X1+2X2 =250 and X1=40,
so 2X2=50, X2 = 25
Point d: (X1= 40, X2= 25)
Point e: Can be calculated by solving the two constrains’ equations as
follows:
5X1+2X2 = 250
-1X1-2X2 =-150 ( multiplied by -1)
4X1 + 0 = 100, so X1 = 25
Substituting in the first equation to get X2 value;
5(25)+2X2=250, 2X2 =125, X2 = 62.5
Point d: (X1= 25, X2= 62.5)
Finding the profit:
Point a: 2(0)+3(75)= $225
Point b: 2(0)+3(0)= $0
Point c: 2(40)+3(0)= $80
Point d: 2(40)+3(25)= $155
Point e: 2(25)+3(62.5)= $237.5
So point e is the answer since this corner pint gives the maximum profit and
X1 should be chosen to be 25 and X2 should be 62.5 to maximize the profit.
Problem 9.8:
The objective function is: Maximize P= 4X1+4X2
Using excel to solve the problem
By drawing parallel lines in every assumed profit and I started by $50 and I
end up with $148 which is the maximum profit and corner point is applied
below to confirm;
Point a: (0,0) P = 4(0) + 4(0) = 0
Point b: (0,30) P = 4(0) + 4(30) = 120
Point c: (18.5,18.5) P = 4(18.5) + 4(18.5) = 148
Point d: (30,0) P = 4(30) + 4(0) = 120
The optimal solution will be at point C (X1=18.5, X2=18.5), P= 148
Problem 9.9:
Let: X1= Brand X.
X2= Brand Z.
Objective function: Minimize cost = 2X1+3X2
1st constraint: 5 X1+10X2 ≥ 90
2nd constraint: 4 X1+3 X2 ≥ 48
3rd constraint: 0.5X1+0 X2 ≥ 1.5
4th and 5th constrains: X1 ≥ 0, X2 ≥ 0
Problem 9.10:
Applying the corner point method to locate the optimum answer;
Point a: solving the 2nd and 3rd constrains equations as follows;
From 3rd constraint’s equation X1=3, so substituting 2nd constraint’s
equation; 3X2=36, X2= 12.
Point b: solving the 1st and 2nd constrains equations as follows
5X1+10X2 = 90
-5X1-3.75X2 = -60 ( multiplied by –(5/4)
0 + 6.25X2 = 30, so X2 = 4.8
Substituting in constrain 1, X1= 8.4
Point c:
We have X1 is 18, so X2= 0
Finding the cost:
Point a: 2(3)+3(12)= $42
Point b: 2(8.4)+3(4.8)= $31.2
Point c: 2(18)+3(0)= $36
So point b is the answer since this corner pint gives the minimum cost and
X1 should be chosen to be 8.4 and X2 should be 4.8 to minimize the cost.
Problem 9.11:
Let:
X1 = pounds of compost
X2 = pounds of sewage
Objective function: Minimize cost = 5X1 + 4X2
1st constraint: X1 + X2 ≥ 60
2nd constraint: X1 ≥ 30
3rd constraint: X2 ≤ 40
4th and 5th constrains: X1 ≥ 0, X2 ≥ 0
Point a: (X1=30, X2=40),
The cost is (5)(30)+(4)(40)= 310
Point b: X1=30, substitute in X1+X2=60, X2=30,so the corner point is
(X1= 30, X2= 30)
The cost is (5)(30)+(4)(30)= 370
Point a, gives the minimized cost which is 310 and it is the optimal answer.
Problem 9.12:
Objective function: Minimize cost = 1X1 + 2X2
1st constraint: 1X1 + 3X2 ≥ 90
2nd constraint: 5X1+1X2 ≥ 100
3rd constraint: 3X1+2X2 ≥ 120
Using QM for windows to solve the problem;
The cost of 80 is closer to the optimum cost, by using the iso-cost method;
Using the corner point method to confirm the exact result:
Point b:
5X1 + X2 = 100
(1)
3X1 + 2X2 = 120 (2)
Multiply equation (1) by -2:
We have -7X1 = -80, X1 = 11.43 substitute in equation (1), X2 = 42.85
,so Point b is (X1 =11.43, X2 =42.85) .The cost = (1)(11.43) + 2(42.85) =
97.1
For point c:
X1 + 3X2 = 90
(1)
3X1 + 2X2 = 120 (2)
Multiply equation (1) by -3:
We have -7X2 = -150, X1 = 21.43 substitute in equation (1), X2 = 25.71
Point c = (X1 =25.71, X2 =21.43)  cost = (25.71) + 2(21.43) = 68.6 Which
is the lowest cost, point c is the answer.