INTENSIVE AND EXTENSIVE FARMING 819 



tiplying the daily requirement by the days in the feeding period, and 

 dividing by 2000, the pounds in a ton. To illustrate: 20 pounds silage 

 X 61 clays, + 35 pounds silage X 175 days, -S- 2000 = 3.672 tons of 

 silage per cow, thus: 



(20 X 61) + (34 X 175) 



2000 



3.672. 



It is next necessary to ascertain the acres of each crop as given in the 

 third and fourth columns of the table. To illustrate, the silage requirement 

 per cow, 3.672 tons, divided by the yield of silage per acre gives the acres 

 of silage corn- required per cow (3.672 -4- 10 = .367 acres). 



Let x equal the number of cows plus one (bull) and the tabulation 

 is as follows: 



Crop. 



Pasture 



Soiling corn 



Silage corn 



Corn for grain 



Oats 



Wheat 



Hay 



Area of farm, 110 acres, equals 



Tons of Feed Required for 



1 Cow or Bull 5 Horses 



4.562 

 4.562 



13 '688 



Acres of Crops Required for 



1 Cow or Bull. 5 Horses. 



l.OOOx + 

 .082x + 

 .367x + 

 .224x + 

 .261x + 

 .261x + 

 .381x + 



2.576x + 



3.259 

 5.704 

 5.704 

 4.562 



19.229 



The equation is solved for the value of x, the number of cows plus 

 one, as follows: 



2.576x = 110 — 19.229 = 90.771 -v- 2.576 = 35 . 24 or 34 . 24 cows and 1 bull. 



With the crop yield assumed and feeding system given the 110 acres 

 will maintain 34 cows, 1 bull and 5 horses, and will provide for an acreage 

 of wheat, as a cash crop, equal to the acreage of oats. 



Substituting the value of x (35.24) in the third column of the above 

 table and adding the acres of the several crops for horses given in the 

 fourth column, the acres of pasture and crops are as follows: 



Crop. Acres. 



Pasture 35 . 240 



Soiling corn 2 . 889 



Silage corn 12.933 J. ! 



Corn for grain 1 1 . 153 



Oats 14 . 900 



Wheat 14 . 900 



Hay 17.988 



Total 110.003 



Pasture 35 



Cultivated crops 75 



