Amit's mathematics teacher has given him three very long lists of problems with the instruction to submit not more than 100 of them (correctly solved) for credit. The problem in the first set are worth 5 points each, those in the second set are worth 4 points each, and those in the third set are worth 6 points each. Amit knows from experience that he requires on the average 3 minutes to solve a 5 point problem, 2 minutes to solve a 4 point problem, and 4 minutes to solve a 6 point problem. Because he has other subjects to worry about, he can not afford to devote more than $$3\dfrac{1}{2}$$ hours altogether to his mathematics assignment. Moreover, the first two sets of problems involve numerical calculations and he knows that he cannot stand more than $$2\dfrac{1}{2}$$ hours work on this type of problem. Under these circumstances, how many problems in each of these categories shall he do in order to get maximum possible credit for his efforts? Formulate this as a LPP.

Asked by Sakshi | 1 year ago |  49

##### Solution :-

Let Amit correctly solves x problems from the first set, y problems from the second set and z problems from the third set.
Given,
Amit cannot submit more than 100 correctly solved problems.

x+y+z≤100

The problem in the first set are worth 5 points each,those in the second set worth 4 points each and those in the third set worth 6 points each.
Therefore, x problems from the first set worth 5x points, y problems from the second set worth 4y points and z problems from the third set worth 6z points.

Thus, total credit points will be

5x+4y+6z5x+4y+6z.

Let Z denotes the total credit of Amit

Z = 5x+4y+6z

It requires 3 minutes to solve a 5 point problem, 2 minutes to solve a 4 point problem and 4 minutes to solve a 6 point problem.Therefore,x problems from the first set require 3x minutes, y problems from the second set require 2y minutes and z problems from the third set require 4z minutes.
Thus, the total time require by Amit will be (3x + 2y + 4z) minutes.
It is given that the total time that Amit can devote on his mathematics assignment is $$3\dfrac{1}{2}$$hours i.e. 210 minutes.312hours i.e. 210 minutes.

3x+2y+4z≤210
Further, it is given that the total time that Amit can devote in solving first two types of problems cannot be more than $$2\dfrac{1}{2}$$ hours​  i.e. 150 minutes150 minutes.

x+2y≤150

Number of problems cannot be negative.Therefore,

x, y≥0x, y≥0

Hence, the required LPP is follows:

Maximize  Z = 5x+4y+6z

subject to
x+y+z≤100

3x+2y+4z≤210

3x+2y≤150x≥0, y≥0

Answered by Aaryan | 1 year ago

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