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//example 2.2
//calculate maximum area that can be irrigated
clc;
//Given
Q=0.0108, //discharge through well
y=0.075, //average depth of flow
I=0.05, //average infiltration rate
A=0.1, //area to cover
Amax=Q/I;
mprintf("Maximum area that can be irrigated =%f hectare.",Amax);
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