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Linear programming — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsLinear programming5 Marks
Question
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 $3\frac{1}{2}$ 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 $2\frac{1}{2}$ 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.
✓

Answer

Given information can be tabulated as below:
Sets
Time requirement
points
I
3
5
II
2
 
III
4
6
Time for all three sets $=3\frac{1}{2}$ hours
Time for set I and II $=2\frac{1}{2}$ hours
Number of quations maximum 100
Given, each question from set I, II, III earn 5,4,6 points respectively, so x questions of set I, y questions of set II and z questions of set III earn 5x, 4y and 6z points, let total point credit be U

So, U = 5x + 4y + 6z

Given, each question of set I, II and III require 3,2 and 4 minutes respectively, so x questions of set I, y questions of set II and z questions of set III require 3x, 2y and 4z mimutes respectively but given that total time to devote in all three sets is

$3\frac{1}{2}$ hours = 210 minutes and first two sets is $2\frac{1}{2}$ hours = 150 minutes

So,

$3\text{x}+2\text{y}+4\text{z}\leq210$ (First constraint)

$3\text{x}+2\text{y}\leq150$ (Second constraint)

Given, total number of questions cannot exceed 100

So, $\text{x}+\text{y}+\text{z}\leq100$ (Third constraint)

Hence, mathematical formulation of LPP is

Find x and y which maximize U = 5x + 4y + 6z

Subject to constraint,

$3\text{x}+2\text{y}+4\text{z}\leq210$

$3\text{x}+2\text{y}\leq150$

$\text{x}+\text{y}+\text{z}\leq100$
$\text{x},\text{y},\text{z}\geq0$

[Since number of questions to solve from each set cannot be less than zero].

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