176 ELEMENTARY TREATISE) ON STOCK FEEDS AND FEEDING 



Use of the Table. Let us make this clearer by computing the 

 standard for a cow weighing 850 Ibs. producing 23 Ibs. of milk 

 daily, testing 5 per cent, butter fat. 



Since 850 is 8.5 X 100, we must multiply our maintenance 

 (0.07 Ib. protein; 0.7 Ib. carbohydrates and o.oi fat) by 8.5. 



0.07 X 8.5 = 0.595 Ib. of protein | Maintenance require- 



o-7 X 8.5 = 5.950 Ibs. of carbohydrates [ mentfora cow weigh- 

 o.oi X ,8-5 = 0.085 Ib. of fat ) ing 850 Ibs. 



Since our cow is producing 23 Ibs. of milk, we divide 23 by 10 

 which gives us 2.3. Multiply the standard for each additional 

 10 Ibs. of 5 per cent, butter fat milk (0.54 Ib. protein, 2.48 Ibs., 

 carbohydrates and 0.18 Ib. fat) by 2.3 which gives us the milk 

 production requirement. 



0.54 



lit 





5 p ce,,t. 



