a toy company manufactures two types of dolls A and B each doll of type B takes twice long as - Maths - Linear Programming

Question

a toy company manufactures two types of dolls A and B each doll
of type B takes twice long as long to produce as one of type A
andthe company would have time to make a maximum 2000 per day if it
produces only type A The supply plastic is sufficientto produce
1500 dolls per day (both A and B combined) Type B requires a fancy
dress of which there are only 600 per day available If the company
makes profit of Rs 3 and Rs 5 per doll respectively on dolls A and
B how many each should be produced per day in order to maximise the
profit? Find the maximum profit

Koka Sri Lakshmi Divya Sai · Accepted Answer

$Let x_{1} dolls of type A and x_{2} dolls of type B ∴ Total profit in Rs.Z=3 x_{1} + 5 x_{2} If t hours are required to produce one doll of type A then the time required to produce one doll of type B will be 2t hours ∴ Total time consumed,per day,to produce x_{1} and x_{2} dolls of type A and type B respectively is x_{1} . t + x_{2} .2 t hrs . But the time available per day is 2000t hrs x_{1} . t + x_{2} .2 t \leq 2000 t or x_{1} + 2 x_{2} \leq 2000 Since plastic is available to produce 1500 dolls only ∴ x_{1} + x_{2} \leq 1500 Also fancy dress is available for 600 dolls per day only ∴ x_{2} \leq 600 Hence,the linear programming problem is as follows Max Z=3 x_{1} + 5 x_{2} Subject to constraints x_{1} + 2 x_{2} \leq 2000 x_{1} + x_{2} \leq 1500 x_{2} \leq 600 and x_{1} \geq 0, x_{2} \geq 0 Sol: Step one: First we consider the constraints as equalities x_{1} + 2 x_{2} \leq 2000 x_{1} + x_{2} \leq 1500 x_{2} \leq 600 Step two: Here we draw lines in two dimensional plane Step three: The shaded region in the fig:is the permissible region for the values of the variables x_{1} and x_{2} Step Four: From the fig it can seen that Z is max. at the point P(1000,500) which is the point of intersection of the lines x_{1} + 2 x_{2} = 2000 x_{1} + x_{2} = 1500 ∴ For Max. Z, x_{1} = 1000 and x_{2} = 500 and Max. Z=Rs. [3 \times 1000 + 5 \times 500] = R s .5500$

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