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Number of bottles of Cartesian Cola Classic produced each day: x>=0
Number of bottles of Cartesian Cola Quantum produced each day: y>=0
Limit of water each day: 100 litres = 100,000 ml
Limit of sugar each day: 100 kg = 100,000 g
Limit of caramel each day: 32 g
Quantity of water required each day:
500x+550y<=100,000.
Dividing the equation by 500:
500x/500+550y/500<=100,000/500
(1) x+1.1y<=200
Quantity of sugar required each day:
600x+200y<=100,000
Dividing the equation by 200:
600x/200+200y/200<=100,000/200
(2) 3x+y<=500
Quantity of caramel required each day:
0.1x+0.2y<=32
Dividing the equation by 0.1
0.1x/0.1+0.2y/0.1<=32/0.1
(3) x+2y<=320
Profit per day:
P=40x+19y
Please see the attached figure
The feasible region is formed by the vertices:
1) (0,0)
2) (0,160)
3) (160/3,400/3)
4) (3500/23,1000/23)
5) (500/3,0)
Evaluating the profit in each vertex:
1) (0,0)=(x,y)→x=0,y=0
P=40(0)+19(0)=0+0→P=0 p
2) (0,160)=(x,y)→x=0, y=160
P=40(0)+19(160)=0+3,040→P=3,040 p
3) (160/3,400/3)=(x,y)→x=160/3, y=400/3
P=40(160/3)+19(400/3)=6,400/3+7,600/3=(6,400+7,600)/3→
P=14,000/3→P=4,666.67 p
4) (3500/23,1000/23)=(x,y)→x=3500/23, y=1000/23
P=40(3500/23)+19(1000/23)=140,000/23+19,000/23
P=159,000/23→P=6,913.04 p
5) (500/3,0)=(x,y)→x=500/3, y=0
P=40(500/3)+19(0)=20,000/3+0
P=20,000/3→P=6,666.67 p
We get a maximum profit of 6,913.04 p when Cartesian Cola produces
152 (x=3,500/23=152.174) bottles of Cartesian Cola Classic and 43 (y=1,000/23=43.478) bottles of Cartesian Cola Quantum
Answer: Cartesian Cola should produce 152 bottles of Cartesian Cola Classic and 43 bottles of Cartesian Cola Quantum each day to maximise their profit.
The profit per day would be 6,913.04 p
Number of bottles of Cartesian Cola Quantum produced each day: y>=0
Limit of water each day: 100 litres = 100,000 ml
Limit of sugar each day: 100 kg = 100,000 g
Limit of caramel each day: 32 g
Quantity of water required each day:
500x+550y<=100,000.
Dividing the equation by 500:
500x/500+550y/500<=100,000/500
(1) x+1.1y<=200
Quantity of sugar required each day:
600x+200y<=100,000
Dividing the equation by 200:
600x/200+200y/200<=100,000/200
(2) 3x+y<=500
Quantity of caramel required each day:
0.1x+0.2y<=32
Dividing the equation by 0.1
0.1x/0.1+0.2y/0.1<=32/0.1
(3) x+2y<=320
Profit per day:
P=40x+19y
Please see the attached figure
The feasible region is formed by the vertices:
1) (0,0)
2) (0,160)
3) (160/3,400/3)
4) (3500/23,1000/23)
5) (500/3,0)
Evaluating the profit in each vertex:
1) (0,0)=(x,y)→x=0,y=0
P=40(0)+19(0)=0+0→P=0 p
2) (0,160)=(x,y)→x=0, y=160
P=40(0)+19(160)=0+3,040→P=3,040 p
3) (160/3,400/3)=(x,y)→x=160/3, y=400/3
P=40(160/3)+19(400/3)=6,400/3+7,600/3=(6,400+7,600)/3→
P=14,000/3→P=4,666.67 p
4) (3500/23,1000/23)=(x,y)→x=3500/23, y=1000/23
P=40(3500/23)+19(1000/23)=140,000/23+19,000/23
P=159,000/23→P=6,913.04 p
5) (500/3,0)=(x,y)→x=500/3, y=0
P=40(500/3)+19(0)=20,000/3+0
P=20,000/3→P=6,666.67 p
We get a maximum profit of 6,913.04 p when Cartesian Cola produces
152 (x=3,500/23=152.174) bottles of Cartesian Cola Classic and 43 (y=1,000/23=43.478) bottles of Cartesian Cola Quantum
Answer: Cartesian Cola should produce 152 bottles of Cartesian Cola Classic and 43 bottles of Cartesian Cola Quantum each day to maximise their profit.
The profit per day would be 6,913.04 p