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?
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?
Let the company manufactures x dolls of type A and y dolls of type B, then
x≥0,y≥0 ..(i)x+y≤1200 ..(ii)y≤x/2⇔x−2y≥ ..(iii)x≤3y+600⇔x−3y≤600 ..(iv)Firstly, draw the graph of the line x+y=1200x12000y01200Putting (0, 0)in the inequality x+y≥1200,we have0+0≤1200⇒ 0≤1200 (which is true)So, the half plane is towards the origin.Secondly, draw the graph of the line x−2y=0x02y01Putting (0, 0)in the inequality x−2y≥0,we have200−2×0≥0⇒ 200≥0 (which is true)So, the half plane is towards the X−axis.Thirdly, draw the graph of the line x−3y=600x6000y0−200Putting (0, 0)in the inequality x−3y≤600,we have 0+3×0≤600⇒0≤600 (which is true)So, the half plane is towards the x−axis.
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)16000→Maximum
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.
Let the company manufactures x dolls of type A and y dolls of type B, then
x≥0,y≥0 ..(i)x+y≤1200 ..(ii)y≤x/2⇔x−2y≥ ..(iii)x≤3y+600⇔x−3y≤600 ..(iv)Firstly, draw the graph of the line x+y=1200x12000y01200Putting (0, 0)in the inequality x+y≥1200,we have0+0≤1200⇒ 0≤1200 (which is true)So, the half plane is towards the origin.Secondly, draw the graph of the line x−2y=0x02y01Putting (0, 0)in the inequality x−2y≥0,we have200−2×0≥0⇒ 200≥0 (which is true)So, the half plane is towards the X−axis.Thirdly, draw the graph of the line x−3y=600x6000y0−200Putting (0, 0)in the inequality x−3y≤600,we have 0+3×0≤600⇒0≤600 (which is true)So, the half plane is towards the x−axis.
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)16000→Maximum
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.
