Question

A car company produces 3 models, model A / B / C. Long-term projections indicate an expected demand of at least 100 model A cars, 80 model B cars and 120 model C cars each day. Because of limitations on production capacity, no more than 200 model A cars and 170 model B cars and 150 model C cars can be made daily. To satisfy a shipping contract, a total of at least 300 cars much be shipped each day. If each model A car sold results in a $1500 loss, but each model B car produces a $3800 profit, each model C car produces a $2500 profit, how many of each type should be made daily to maximize net profits?

Answer 1

model A

model B

model C

the demand of at least 100 model A cars, 80 model B cars and 120 model C cars each day

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no more than 200 model A cars and 170 model B cars and 150 model C cars can be made daily.

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a total of at least 300 cars much be shipped each day.

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each model A car sold results in a $1500 loss, but each model B car produces a $3800 profit, each model C car produces a $2500 profit,

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so our objective function is

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our system is  

subject to

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After introducing slack,surplus,artificial variables

subject to

Iteration-1		Cj	-1500	3800	2500	0	0	0	0	0	0	0	-M	-M	-M	-M
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	S5	S6	S7	A1	A2	A3	A4	MinRatio XB/x2
A1	-M	300	1	1	1	-1	0	0	0	0	0	0	1	0	0	0	300/1=300
S2	0	200	1	0	0	0	1	0	0	0	0	0	0	0	0	0	---
S3	0	170	0	1	0	0	0	1	0	0	0	0	0	0	0	0	170/1=170
S4	0	150	0	0	1	0	0	0	1	0	0	0	0	0	0	0	---
A2	-M	100	1	0	0	0	0	0	0	-1	0	0	0	1	0	0	---
A3	-M	80	0	(1)	0	0	0	0	0	0	-1	0	0	0	1	0	80/1=80→
A4	-M	120	0	0	1	0	0	0	0	0	0	-1	0	0	0	1	---
Z=-600M		Zj	-2M	-2M	-2M	M	0	0	0	M	M	M	-M	-M	-M	-M
		Zj-Cj	-2M+1500	-2M-3800↑	-2M-2500	M	0	0	0	M	M	M	0	0	0	0

Negative minimum Zj-Cj is -2M-3800 and its column index is 2

Minimum ratio is 80 and its row index is 6

The pivot element is 1.

Entering =x2, Departing =A3

Iteration-2		Cj	-1500	3800	2500	0	0	0	0	0	0	0	-M	-M	-M
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	S5	S6	S7	A1	A2	A4	MinRatio XB/x3
A1	-M	220	1	0	1	-1	0	0	0	0	1	0	1	0	0	220/1=220
S2	0	200	1	0	0	0	1	0	0	0	0	0	0	0	0	---
S3	0	90	0	0	0	0	0	1	0	0	1	0	0	0	0	---
S4	0	150	0	0	1	0	0	0	1	0	0	0	0	0	0	150/1=150
A2	-M	100	1	0	0	0	0	0	0	-1	0	0	0	1	0	---
x2	3800	80	0	1	0	0	0	0	0	0	-1	0	0	0	0	---
A4	-M	120	0	0	(1)	0	0	0	0	0	0	-1	0	0	1	120/1=120→
Z=-440M+304000		Zj	-2M	3800	-2M	M	0	0	0	M	-M-3800	M	-M	-M	-M
		Zj-Cj	-2M+1500	0	-2M-2500↑	M	0	0	0	M	-M-3800	M	0	0	0

Negative minimum Zj-Cj is -2M-2500 and its column index is 3.

Minimum ratio is 120 and its row index is 7.

The pivot element is 1.

Entering =x3, Departing =A4,

Iteration-3		Cj	-1500	3800	2500	0	0	0	0	0	0	0	-M	-M
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	S5	S6	S7	A1	A2	MinRatio XB/x1
A1	-M	100	(1)	0	0	-1	0	0	0	0	1	1	1	0	100/1=100→
S2	0	200	1	0	0	0	1	0	0	0	0	0	0	0	200/1=200
S3	0	90	0	0	0	0	0	1	0	0	1	0	0	0	---
S4	0	30	0	0	0	0	0	0	1	0	0	1	0	0	---
A2	-M	100	1	0	0	0	0	0	0	-1	0	0	0	1	100/1=100
x2	3800	80	0	1	0	0	0	0	0	0	-1	0	0	0	---
x3	2500	120	0	0	1	0	0	0	0	0	0	-1	0	0	---
Z=-200M+604000		Zj	-2M	3800	2500	M	0	0	0	M	-M-3800	-M-2500	-M	-M
		Zj-Cj	-2M+1500↑	0	0	M	0	0	0	M	-M-3800	-M-2500	0	0

Negative minimum Zj-Cj is -2M+1500 and its column index is 1

Minimum ratio is 100 and its row index is 1.

The pivot element is 1.

Entering =x1, Departing =A1,

Iteration-4		Cj	-1500	3800	2500	0	0	0	0	0	0	0	-M
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	S5	S6	S7	A2	MinRatio XB/S1
x1	-1500	100	1	0	0	-1	0	0	0	0	1	1	0	---
S2	0	100	0	0	0	1	1	0	0	0	-1	-1	0	100/1=100
S3	0	90	0	0	0	0	0	1	0	0	1	0	0	---
S4	0	30	0	0	0	0	0	0	1	0	0	1	0	---
A2	-M	0	0	0	0	(1)	0	0	0	-1	-1	-1	1	0/1=0→
x2	3800	80	0	1	0	0	0	0	0	0	-1	0	0	---
x3	2500	120	0	0	1	0	0	0	0	0	0	-1	0	---
Z=454000		Zj	-1500	3800	2500	-M+1500	0	0	0	M	M-5300	M-4000	-M
		Zj-Cj	0	0	0	-M+1500↑	0	0	0	M	M-5300	M-4000	0

Negative minimum Zj-Cj is -M+1500 and its column index is 4.

Minimum ratio is 0 and its row index is 5.

The pivot element is 1.

Entering =S1, Departing =A2,

Iteration-5		Cj	-1500	3800	2500	0	0	0	0	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	S5	S6	S7	MinRatio XB/S6
x1	-1500	100	1	0	0	0	0	0	0	-1	0	0	---
S2	0	100	0	0	0	0	1	0	0	1	0	0	---
S3	0	90	0	0	0	0	0	1	0	0	(1)	0	90/1=90→
S4	0	30	0	0	0	0	0	0	1	0	0	1	---
S1	0	0	0	0	0	1	0	0	0	-1	-1	-1	---
x2	3800	80	0	1	0	0	0	0	0	0	-1	0	---
x3	2500	120	0	0	1	0	0	0	0	0	0	-1	---
Z=454000		Zj	-1500	3800	2500	0	0	0	0	1500	-3800	-2500
		Zj-Cj	0	0	0	0	0	0	0	1500	-3800↑	-2500

Negative minimum Zj-Cj is -3800 and its column index is 9

Minimum ratio is 90 and its row index is 3.

The pivot element is 1.

Entering =S6, Departing =S3,

Iteration-6		Cj	-1500	3800	2500	0	0	0	0	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	S5	S6	S7	MinRatio XB/S7
x1	-1500	100	1	0	0	0	0	0	0	-1	0	0	---
S2	0	100	0	0	0	0	1	0	0	1	0	0	---
S6	0	90	0	0	0	0	0	1	0	0	1	0	---
S4	0	30	0	0	0	0	0	0	1	0	0	(1)	30/1=30→
S1	0	90	0	0	0	1	0	1	0	-1	0	-1	---
x2	3800	170	0	1	0	0	0	1	0	0	0	0	---
x3	2500	120	0	0	1	0	0	0	0	0	0	-1	---
Z=796000		Zj	-1500	3800	2500	0	0	3800	0	1500	0	-2500
		Zj-Cj	0	0	0	0	0	3800	0	1500	0	-2500↑

Negative minimum Zj-Cj is -2500 and its column index is 10.

Minimum ratio is 30 and its row index is 4

The pivot element is 1.

Entering =S7, Departing =S4

Iteration-7		Cj	-1500	3800	2500	0	0	0	0	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	S4	S5	S6	S7	MinRatio
x1	-1500	100	1	0	0	0	0	0	0	-1	0	0
S2	0	100	0	0	0	0	1	0	0	1	0	0
S6	0	90	0	0	0	0	0	1	0	0	1	0
S7	0	30	0	0	0	0	0	0	1	0	0	1
S1	0	120	0	0	0	1	0	1	1	-1	0	0
x2	3800	170	0	1	0	0	0	1	0	0	0	0
x3	2500	150	0	0	1	0	0	0	1	0	0	0
Z=871000		Zj	-1500	3800	2500	0	0	3800	2500	1500	0	0
		Zj-Cj	0	0	0	0	0	3800	2500	1500	0	0

Since all  

Hence, optimal solution is arrived

.

.

Model A = 100

Model B = 170

Model C = 150

and maximum profit is $871000

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.

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Hint to avoid too much calculation.

here each model A car sold results in a $1500 loss. so we have to produce the lowest quantity of model A.

but in question mention that minimum quantity for model A is 100, so the first answer is model A = 100 cars

.

now as we can see a profit of model B is higher than the profit of model C

so we have to produce the maximum quantity of model B, but in question mention that no more than 170 model B cars

so the second answer is model B = 170 cars

.

and in question mention that no more than 150 model C cars

so the third answer is model C = 150 cars

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and maximum profit is

Z=-1500(100)+3800(170)+2500(150)

Z=-15000+646000+375000

Z=871000...............that is the maximum profit