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Up to now we have only been thinking about integer solutions. Our farmer uses a whole number of bags of each type of fertilizer per hectare. Many real world problems are like this, but in many situations there is no need to restrict the solution to whole numbers. Consider the manufacturer of the high-grade fertilizer. They don't buy their raw materials by the bag; they buy it by the cubic metre. It is quite likely that their manufacturing formula involves, say, 12.4 cubic metres of crushed phosphate rock together with 5.1 cubic metres of dolomite (or whatever it is they put in fertilizer).
Assume that the farmer does not need to use a whole number of bags on each hectare. Which of the following is a feasible solution to the fertilizer problem? Move the red dot around to see which are within the feasible region.
i) 3.4 bags of high grade and 3.6 bags of low
grade?
ii) 1.3 bag of high grade and 11.0 bags of low grade?
iii) 3.8 bags of high grade and 7.8 bags of low grade?
iv) 4.5 bags of high grade and no low grade?
v) 2.2 bags of high grade and 8.4 bags of low grade?
Use the applet to determine which of the given combinations of x and y are in the feasible region for the following linear inequalities. (You may assume non-negativity constraints for x and y.)
a) 3x+5y≤35
i) x=1.4, y=5.4?
ii) x=9.5, y=1.5?
iii) x=2.7, y=7.2?
b) 4x+6y≤54
i) x=1.4, y=5.4?
ii) x=9.5, y=1.5?
iii) x=2.7, y=7.2?
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