Question

A) Solve the following linear program graphically. How many extreme points exist for this problem?

​

Min	150X + 210Y
s.t.	3.8X + 1.2Y ≥ 22.8
	Y ≥ 6
	Y ≤ 15
	45X +  30Y = 630
	X, Y ≥ 0

B)

Comfort Plus Inc. (CPI) manufactures a standard dining chair used in restaurants. The demand forecasts for quarter 1 (January–March) and quarter 2 (April–June) are 3700 chairs and 4200 chairs, respectively. CPI has a policy of satisfying all demand in the quarter in which it occurs.

The chair contains an upholstered seat that can be produced by CPI or purchased from DAP, a subcontractor.  DAP currently charges $12.50 per seat, but has announced a new price of $13.75, effective April 1.  CPI can produce the seat at a cost of $10.25.   CPI can produce up to 3800 seats per quarter.

​

Seats that are produced or purchased in quarter 1 and used to satisfy demand in quarter 2 cost CPI $1.50 each to hold in inventory, but maximum inventory cannot exceed 300 seats.

​

Formulate this problem as a linear programming problem.

Answer 1

A min Z=150x+ 210X st 3-6X + 1-27 > 22-9 YS15 45 X+307 =530 X/Y > To draw constraint 3.8*+1:27>22:8 Treat it as 3.8x+12y= 228 y =19 24 50 368000+ 20y)=228 Yox=6 and usx+ 30y =636 X=0 Y = 21 .Y=0 > x=14 praw the graph 45x+30 (15)=630 x=4 45x+ 3006)=630 *=10. Corigley (415) Y-15 YEG Scanned Calma Scannel

Extieme points 0 4915) O (10) obj function value 15004+ 210 (15)= 3750 150610 † 210C6)= 2760 and mis value is occurrs at (10,6 Erosios = 2760. Scanned with CamScanner