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Question

# A manufacturer produces nuts and bolts. It takes 1 h of work on machine A and 3 h on machine B to produce a package of nuts. It takes 3 h on machine A and 1 h on maching B to produce a package of bolts. He earns a profit of Rs. 17.50 per package on nuts and Rs. 7.00 per package on bolts. How many package of each should be produced each day so as to maximize his profit, if he operates his machines for at the most 12 h a day ?

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Solution

## Let the manufacture produces x nuts and y bolts. We construct the following table : ItemNumberofMachine A (in h)Machine B (in h)Profit(in Rs.)packagesNutsx1x3x17.50xBoltsy3y1y7yTotalx+yx+3y3x+y17.50x+7yAvailability1212 ∴ Total profit, Z = 17.5x + 7y i.e., maximize Z = 17.5 x + 7y ........(i) Subject to constraints x+3y≤12 .........(ii) 3x+y≤12 ..........(iii) x≥0, y≥0 ..........(iv) Firstly, draw the graph of the lines x + 3y = 12 x012y40 Putting (0, 0) in the inequality x + 3y ≤ 12, we have 0+3×0≤12 ⇒0≤12 (which is true) So, the half plane is towards from the origin. Since, x,y≥0 So, the feasible region lies in first quadrant. Secondly, draw the graph of the line 3x+y=12 x40y012 Putting (0, 0) in the inequality 3x+y≤12, We have 3×0+0≤12 ⇒0≤12 (which is true) So, the half plane is towards from the origin. On solving equations x+3y=12 and 3x+y=12, we get B(3, 3). ∴ Feasible region is OABCO. The corner points of the feasible region are O(0, 0), A(4, 0), B(3, 3) and C(0, 4). The values of Z at these points are as follows: Corner pointZ=20x+10yO(0, 0)0A(4, 0)70B(3, 3)73.50→MaximumC(0,4)28 The maximum value of Z is Rs. 73.50 at (3, 3). Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit Rs. 73.50

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