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Linear programming problems are related to distribution of resources which are limited in the best way possible so as to minimise costs or maximise profits
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A linear programming model can be formulated by following the steps given below:
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Formulating a mathematical model for a situation based on the given constraints and presenting it graphically
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- A mathematical model consisting of constraints or objective functions can be obtained from the situation or problem given
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The model can be formulated by using the variables \(x \text{ and } y\) with the constraints in each situation being \(\leqslant, \ \geqslant, \ \text{< or >} \)
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Given a straight line \(ax+by=c\), then,
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| Region above the straight line |
Region below the straight line |
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Satisfies the inequalities
\(ax+by \geqslant c \text{ and } ax+by \text{ > }c\)
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Satisfies the inequalities
\(ax+by \leqslant c \text{ and } ax+by \text{ < }c\)
where \(b \text{ > }0\)
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| Region on the right side of line |
Region on the left side of line |
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Satisfies the inequalities
\(ax \geqslant c \text{ and } ax \text{ > }c\)
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Satisfies the inequalities
\(ax \leqslant c \text{ and } ax \text{ < }c\)
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\(\geqslant \text{ or } \leqslant\), then a solid line is used |
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\(\text{< or >}\), then a dotted line is used |
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| Example |
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Write a mathematical model for the following situation:
The perimeter of the rectangular photo frame must not be more than \(180 \text{ cm}\).
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Present the inequalities \(x-2y \geqslant -4\) graphically. |
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| Solution: |
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Supposed \(x \text{ and } y\) 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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Optimisation in linear programming
- Note that \(ax +by\) is a linear expression
- In general, an objective function is written as
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| Remark |
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Steps to determine the suitable value of \(k\) for
\(k=ax+by\):
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Note that \(a\text{ and }b\) are coefficients of \(x\text{ and }y\) respectively
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Find the common multiples of \(a\text{ and }b\)
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Take \(k\) as the common multiple
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Example:
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| Example |
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The diagram above shows the shaded region that satisfies a few constraints of a situation.
Given \(k = x+2y\), find
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the maximum value of \(k\),
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the minimum value of \(k\). |
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| Solution: |
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Substitute the maximum point for the shaded region, which is \((15,55)\) into \(k = x+2y\).
\(\begin{aligned} k&=15+2(55)\\ &=125 \end{aligned}\)
Therefore, the maximum value of \(k\) is \(125\).
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Substitute the minimum point for the shaded region, which is \((15,8)\) into \(k = x+2y\).
\(\begin{aligned} k&=15+2(8)\\ &=31 \end{aligned}\)
Therefore, the minimum value of \(k\) is \(31\)
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