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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
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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.
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
Let Z denotes the total credit of Amit
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 \[3\frac{1}{2} \text{ hours i . e . 210 minutes } .\]
Further, it is given that the total time that Amit can devote in solving first two types of problems cannot be more than
Maximize
\[3x + 2y + 4z \leq 210\]
\[ 3x + 2y \leq 150\]
\[x \geq 0, y \geq 0\]
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Minimize Z = 4x + 5y
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| Inequations | Equations | X intercept | Y intercept | Region |
| 5x + y ≥ 10 | 5x + y = 10 | ( ___, 0) | (0, 10) | Away from origin |
| x + y ≥ 6 | x + y = 6 | (6, 0) | (0, ___ ) | Away from origin |
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∵ Origin has not satisfied the inequations.
∴ Solution of the inequations is away from origin.
The feasible region is unbounded area which is satisfied by all constraints.
In the figure, ABCD represents
The set of the feasible solution where
A(12, 0), B( ___, ___ ), C ( ___, ___ ) and D(0, 10).
The coordinates of B are obtained by solving equations
x + 4y = 12 and x + y = 6
The coordinates of C are obtained by solving equations
5x + y = 10 and x + y = 6
Hence the optimum solution lies at the extreme points.
The optimal solution is in the following table:
| Point | Coordinates | Z = 4x + 5y | Values | Remark |
| A | (12, 0) | 4(12) + 5(0) | 48 | |
| B | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | ______ |
| C | ( ___, ___ ) | 4( ___) + 5(___ ) | ______ | |
| D | (0, 10) | 4(0) + 5(10) | 50 |
∴ Z is minimum at ___ ( ___, ___ ) with the value ___
If z = 200x + 500y .....(i)
Subject to the constraints:
x + 2y ≥ 10 .......(ii)
3x + 4y ≤ 24 ......(iii)
x, 0, y ≥ 0 ......(iv)
At which point minimum value of Z is attained.
