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A toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
 


Solution

Let the company manufactures x dolls of type A and y dolls of type B, then

x0,y0                           ..(i)x+y1200                         ..(ii)yx/2x2y              ..(iii)x3y+600x3y600    ..(iv)Firstly, draw the graph of the line x+y=1200x12000y01200Putting (0, 0)in the inequality x+y1200,we have0+01200 01200    (which is true)So, the half plane is towards the origin.Secondly, draw the graph of the line x2y=0x02y01Putting (0, 0)in the inequality x2y0,we have2002×00 2000    (which is true)So, the half plane is towards the Xaxis.Thirdly, draw the graph of the line x3y=600x6000y0200Putting (0, 0)in the inequality x3y600,we have 0+3×06000600 (which is true)So, the half plane is towards the xaxis.

Since, x, y 0

So, the feasible region lies in the first quadrant. The point of x = xy intersection of lines x - 3y = 600 and x + y = 1200 is B(1050, 150); of lines x = 2y and x + y = 1200 is C (800, 400)

Let Z be the total profit, then

Z =  12x + 16y

Feasible region is OABCO.

The corner points of the feasible region are A(600, 0), B(1050, 150) and C(800, 400). The values of Z at these points are as follows:

Corner pointZ=12x+16yA(600, 0)7200B(1050, 150)15000C(800, 400)16000Maximum

The maximum value of Z is 16000 at C(800, 400).

Thus, 800 and 400 dolls of types A and type B should be produced respectively to get the maximum profit of Rs 16000.

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