Alex’s Furniture Warehouse Linear Programming for a Furniture Company

Our purpose:  To maximize our profit

Variables:  X= The amount of tables  Y= the amount of chairs

Constraints:  It takes 4 hours to make a table, and 2 hours to make a chair. There are 336 hours available (2 weeks).  We must cannot make a negative amount of chairs or tables.  A table is 19.5’ by 4’,or 78 square feet. A chair is 2.5’ by 3’, or 7.5 square feet. There are only 5,000 square feet available in the warehouse.  We can make, at most, 100 chairs.

Cost Function:  A table can be sold at $500.  A chair can be sold at $250.  f(x,y)=500x+250y

Equations of Constraints:  X≥0  Y≥0  4x+2y≤ 336  78x+7.5y≤ 5,000  Y≤100

Points: X≥0 Y≥0 78x+7.5y≤5,000 X=0 Y=0 (0,0) 78x+7.5y=5, y=-78x+5,000 4x+2y≤336 Y=-10.4x x+2y=336 Y=-10.4(0) y=-4x+336 Y= (0,666.67) Y=-2x+168 Y=-10.4x Y=-2(0)+168 0=-10.4x Y=168 (0,168) =-10.4x Y=-2x+168 X=64.1 (64.1,0) 0=-2x =-2x X=84 (84,0)

Vertices: (0,0) (0,100) (64.1,0) (already calculated as points) 1) 4x+2y≤3361) Y=-2(59.365)+1682) 2y=100 4x+2y=336Y= – 4x+2y=336 2y=-4x+336Y= x=-136 Y=-2x+168(59.4,49.3)X=34 Y=100 1) 78x+7.5y≤5,0002) y≤100(34,100) 78x+7.5y=5,000y=100 78x+7.5(-2x+168)=5,0004x+2y≤336 78x-15x+1260=5,0004x+2y=336Final Vertices: 63x=37402(y)=(100)2(0,0) (0,100) (64.1,0) X=59.365(59.4,49.3) (34,100)

Our Graph :

Solutions: Cost Function: f(x,y)=500x+250y Vertex: (0,0)Vertex: (59.4,49.3) f(0,0)=500(0)+250(0)=0 Because we can’t make a fraction of a chair/table, we must round: (59,49) Vertex: (0,100)f(59,49)=500(59)+250(49)= f(0,100)=500(0)+250(100)=29,500+12,250=41, ,000=25,000 Vertex: (64.1,0)Vertex: (34,100) Because we can’t make a fraction f(34,100)=500(34)+250(100)= Of a chair/table, we must round: (64,0)17,000+25,000=42,000 f(64,0)=500(64)+250(0)= 32,000+0=32,000

Conclusion: In conclusion, the best solution is $41,750 at (59,49). This means that we will make the most profit ($41,750) when we make and sell 59 tables, and 49 chairs. Because our goal was to maximize our profit, when we put the various vertices into our cost function, we wanted our answer to be the largest number, which simulates the largest profit we can make. Finally, we know that this is the best solution because we put all the vertices into our cost function, and (59,49) came out with the largest profit.