The nutrients needed by plants are mostly taken from the soil. If the supply of nutrients in the soil is abundant, crops will grow well and produce high yields. However, if one of the nutrients needed is in short supply, plant growth is limited and crop yields are reduced. Nutrient soil test and fertilizers application is important to supplement nutrients needed by crops. More and better quality food and cash crops can be produced and the fertility of soils, which have been over-exploited, can also be restored.

Soil test laboratories provide a critical step in management decisions of nutrients for optimum crop production. A Good Nutrient Soil Test is able to predict nutrient application rate for responsive soil.

__How to interpret fertilizer recommendation: __

__Example 1__

After soil test, a farmer is advised that** a maize crop will require 80 kg N, and 60 kg P_{2}O_{5} to achieve 7 tons per Ha.**

**The available fertilizers in the market are: DAP (18:46:0) and CAN (26:0:0).**

**The farmer comes to you for assistance and ask “How much of each fertilizer will be required for optimum maize yields?”**

**Method:**

__Step 1: __

**4R Plant Nutrient Management – Rate, Source, Time & Place.**

**Apply the basal fertilizer first i.e. DAP to boost root growth (P _{2}O_{5})**

**Determine how many kg DAP are required to supply 60kg P _{2}O_{5}? **

100 kg DAP supplies 46 kg P_{2}O_{5}

Y 60 kg P_{2}O_{5 }per Ha

** **

**Y = ** __60 kg P _{2}O_{5 }/ha x __

__100 kg DAP__

46 kg P_{2}O_{5}

_{ }

**Y ** = __60 kg P _{2}O_{5} /ha x __

__100 kg DAP__=

__60 x 100 kg DAP per Ha__

46 kg P_{2}O_{5} 46

= **130.4 kg DAP per Ha**

** **

__Step 2__: Determine how much nitrogen is supplied by 130.4 kg DAP

100 kg DAP supplies 18 kg N

130.4 kg DAP per Ha Y kg N

Y = __130.4 kg DAP / ha __x 18 kg N

100 kg DAP

** **

**Y = 23.47 kg N /ha**

__Step 3__: Determine how much additional nitrogen required from CAN

80 N kg/ha required – 23.47 kg N per Ha supplied by DAP = 56.53 kg N from CAN

100 kg CAN 26 kg N

Z kg CAN 56.53 kg N

Z = 56.53 kg N/ha x 100 kg CAN per Ha

^{——————————————————–}

^{ }26 kg N per Ha

^{ }

**Z = 217.4 kg CAN per Ha.**

** **

**Conclusion: **

Based on the above example where we are assuming our soil test recommend that a maize crop will require 80 kg N, and 60 kg P_{2}O_{5} to achieve 7 tons per Ha. We can conclude that:

**How much of each fertilizer is required for optimum maize yields = 130 kg DAP and 4 (50kg bags of CAN)****To arrive at bags Divide Kg of Each by Capacity of Bag e.g. 50 for 50 kg bag****How much per plant: Convert kgs of fertilizer into grams then divide by plant population**

** **

__Example 2__

**A maize crop requires that, there is 80kg of Nitrogen(N), and 60kg of Phosphorus(P) to achieve 7 tons per Ha**

**Only Triple super phosphate (TSP) (0:46:0) fertilizer, CAN (26:0:0) and urea (46:0:0) are available.**

**How ****much of each fertilizer is required for optimum maize yields.**

__Step 1: __Determine how many kg of P_{2}O_{5} is equivalent to P_{2}

**Atomic weight of P _{2}O_{5 }is [2 x 31.0g (P) =62g] + [5 x 16.0g (O) =80] **

** Total =62.0 +80 =142g**

** 62g P in**** 142g P _{2}O_{5}**

__Step 2: __Calculate the P_{2}O_{5 }equivalent to 60kg P by calculating the conversion factor

** If 62g P in **** 142g P _{2}O_{5, }conversion factor for P_{2}O_{5} into P is: **

** 142/62 =2.29 **

**(Implying you can get P by dividing with this factor or get P _{2}O_{5} by multiplying with this factor)**

__Step 3: __How much P_{2}O_{5 }will supply 60kg P

** 142g P _{2}O_{5} 62g P **

** X 60kg P**

**X = 60 kg P x 142kg P_{2}O_{5}**

** 62kg P**

** X = 137.4 kg P _{2}O_{5}**

**Step 4: How much TSP will supply 137.4 kg P _{2}O_{5}? **

100 kg TSP 46 kg **P _{2}O_{5}**

Y kg TSP **137.4 kg P _{2}O_{5}**

**Y = 137.4 kg P_{2}O_{5 }X **

__100 kg TSP__

46 kg **P _{2}O_{5}**

**Y = 298.7 kg TSP**

** **

**Step 5: How much CAN is required to supply 80kg N? **

**100 kg CAN 26 kg N**

**Z kg CAN 80 kg N**

**Z = 80 kg N/ha x 100 kg CAN**

** 26 kg N**

**Z = 307.7 kg CAN per Ha**

** **

**Assignment: if the farmer decides to use Urea instead of CAN, How much urea will be required to supply 80kg N? **

**Recommended nutrient rate 80 kg N/ha; N % of Urea is 46%. How many kg of Urea do you need in a hectare? **

__80 kg N/ha x 100 kg Urea__

** 46 kg N**

** = 173.9 kg Urea per Ha **