7.1 |
Linear Programming Model |
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Forming a Mathematical Model for a Situation Based on Given Constraints and Representing the Model Graphically |
A mathematical model can be formulated by using the variables \(x\) and \(y\) with the constraints in each situation being \(\leq\), \(\ge\), \(\lt\) or \(\gt\).
- The region above the straight line \(ax+by=c\) satisfies the inequalities \(ax+by\ge c\) and \(ax+by\gt c\), where \(b \gt 0\).
- The region below the straight line \(ax+by=c\) satisfies the inequalities \(ax+by\le c\) and \(ax+by \lt c\), where \(b\gt 0\).
- The region on the right side of the line \(ax=c\) satisfies the inequalities \(ax\ge c\) and \(ax \gt c\).
- The region on the left side of the line \(ax=c\) satisfies the inequalities \(ax\le c\) and \(ax \lt c\).
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If a mathematical model involves signs like:
- \(\ge\) or \(\le\), then a solid line \((\overline{\hspace{1cm}})\) is used.
- \(\lt\) or \(\gt\), then a dotted line \((\text{-} \text{ -} \text{ -} \text{ -}\text{ -})\) is used.
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Objective Function |
An objective function is written as:
\(k=ax+by\)
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Example \(1\) |
(a) |
Write a mathematical model for the following situation: |
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The perimeter of the rectangular photo frame must not be more than \(180\) cm. |
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(b) |
Present the inequalities \(x-2y \geqslant -4\) graphically. |
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Supposed \(x\) and \(y\) are the width and length of the rectangular photo frame.
Then, \(2x+2y \text{ < }180\).
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Given \(x-2y \geqslant -4\).
Since \(b=-2 \ (\text{< }0)\).
Hence, the region lies below the line \(x-2y=-4\).

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Example \(2\) |

The diagram above shows the shaded region that satisfies a few constraints of a situation.
Given \(k = x+2y\), find
(a) |
the maximum value of \(k\), |
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the minimum value of \(k\). |
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(a) |
Substitute the maximum point for the shaded region, which is \((15,55)\) into \(k = x+2y\). |
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\(\begin{aligned} k&=15+2(55)\\ &=125. \end{aligned}\) |
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Therefore, the maximum value of \(k\) is \(125\). |
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(b) |
Substitute the minimum point for the shaded region, which is \((15,8)\) into \(k = x+2y\). |
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\(\begin{aligned} k&=15+2(8)\\ &=31. \end{aligned}\) |
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Therefore, the minimum value of \(k\) is \(31\). |
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