solutiongiven problem MIN Zx=1500x1+2400x2subject to4x1+x2≥242x1+3x2≥42x1+4x2≥36x1≤14x2≤14and x1,x2≥0;−−−−−−−−−−−−−−−−−−−−−−(1). To draw constraint 4x1+x2≥24→(1)Treat it as 4x1+x2=24When x1=0 then x2=?⇒4(0)+x2=24⇒x2=24When x2=0 then x1=?⇒4x1+(0)=24⇒4x1=24=x1=6∣∣x1x2 0 24 6 0∣∣−−−−−−−−−−−−−−−−−−−−−−similar(2).To draw constraint 2x1+3x2≥42→(2)∣∣x1x2 0 14 21 0∣∣−−−−−−−−−−−−−−−−−−−−−−(3)To draw constraint x1+4x2≥36→(3)∣∣x1x2 0 9 36 0∣∣−−−−−−−−−−−−−−−−−−−−−−(4) To draw constraint x1≤14→(4)∣∣x1x2 14 0 14 1∣∣−−−−−−−−−−−−−−−−−−−−−−(5)To draw constraint x2≤14→(5)∣∣x1x2 0 14 1 14∣∣−−−−−−−−−−−−−−−−−−−−−−now we draw graph of above 1 to 5 data set. graph

yellow line x2≤14brown line x1≤14blue line x1+4x2≥36green line 2x1+3x2≥42red line 4x1+x2≥24−−−−−−−−−−−−−−−−−−−−−−The value of the objective function at each of these extreme points is as follows:∣∣Extreme PointCoordinates(x1,x2) A(2.5,14)B(14,14)C(14,5.5)D(12,6)E(3,12) Objective function valueZ=1500x1+2400x2 1500(2.5)+2400(14)=373501500(14)+2400(14)=546001500(14)+2400(5.5)=342001500(12)+2400(6)=324001500(3)+2400(12)=33300∣∣The miniimum value of the objective function Z=32400 occurs at the extreme point (12,6).Hence, the optimal solution to the given LP problem is : x1=12,x2=6 and min Z=32400.
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