Question

40. Furniture. A furniture manufacturing company manufac- tures dining-room tables and chairs. The relevant manufactur- ing data are given in the table below. oitoon Labor-Hours per Unit Maximum Labor-Hours Table Chair Available per Day 8 2 400 Department Assembly Finishing Profit per unit 1200 $25 (A) How many tables and chairs should be manufactured each day to realize a maximum profit? What is the maximum profit? (B) Discuss the effect on the production schedule and the maximum profit if the marketing department of the com- pany decides that the number of chairs produced should be at least four times the number of tables produced.

Answer 1

no bhag A Let Number of table = Number of chairs=y Then given dater in linear Programming problem Mom za gokt 25 8x+ By < 400 dn+ y 5120 ny7,0 Convert into stand and form by adding stack variables 8, ka Man z= gon +254 + 0.3, +0.52 &xt 2y + o.s. -400 o o duty & ke = 120 ky, s, ,52710 & c; I go as min latio By B co XB x y si se x Si O 400 $ 2 1 400 = 50 min Se o 120 2 1 0 1 129 260 zj-cj 1-90 -25 o or Tlmost Negative) Entering Variable ex; Leaving variablets, Pirot Pivot element - 8 (Key element) 90 25 o o Min Ratio & XB 1 BCB no x Is. Sa x 90 50 1 4 78 oso = 200 S o aloo D -V4 1 - 40 - (min) Z-cj 10 -2.5 11.25 of (most Negative)

Va c - ols les Entering variable try; Leaving variable is key element z 12 Igo 25 o oo In y S, S. 1 0 V4 -42 7 25 40 0 1 -12 2 Zj-cj To o 10 st VZj-cj »o opteimal sobition is arrived. x= 40 4240 No. of tables = 40 ; no of Chairs 240 Max prosit = (40 x 90lt 140+25) - $4600 mavinum profit $4600 T The number of chairs produced should be at least the number of tables produced. ie. 4x stay So now Max za goxt 25y + 0. 5, +0. 9. 10.53 8x + 2y & S, a 400 de+ y + 5z = 120 Hery 45320 xy, S., S, S3710 four times noun

min fatis 2XB4 läñ at c; go 25 CB Xo x y o 400 8 2 0 120 2 1 o s 1 0 o u 0 1 o 33 0 0 17 O 3-4 -30 -25 o o o | Ilmost Negative) Batang Jarable = x 'i kending variable : kay Clement =4 Cj 90 B 25 0 0 0 Min Ratio a 9 S. S2 si o 400 o 4 o -2400 = 100 2 o 120 o 1 o 1 -12 120 - 80 (min x 90 oliy o 0 14 10-41.5 o o 22.5 I most negatine) entering Variablety : Leaving variables se key elnet = 3/2 c; go 25 o o o A B CB XB x y s. s. I S. O sol o o I -813 –213 , 25 solo 1 o 213 1/3 & go 20 o 0 % 16 Zj-cj To o o 31.6667 6.66677 & Zj-Lj 70 ; fence optimal solution is arrived. No. of tables - 20 No. of chairs = 80 Maximum profit = (90x20 )+ ( 25480 )= 3 800

When we change our system e. we reakce produces number of chair produced should be at least four times the number of tables produces then po profit decrease by 4600-3800 = $800.