Question

A farmer uses two types of fertilizers. A 50-lb bag of Fertilizer A contains 10 lbs. of nitrogen, 2 lbs. of phosphorus, and 6 lbs. of potassium. A 50-lb bag of Fertilizer B contains 5 lbs. each of nitrogen, phosphorus, and potassium. The minimum requirements for a field are 420 lbs. of nitrogen, 240 lbs. of phosphorus, and 360 lbs. of potassium. If a 50-lb bag of Fertilizer A costs $35 and a 50-lb bag of Fertilizer B costs $25, find the amount of each type of fertilizer the farmer should use to minimize his cost while still meeting the minimum requirements. What is the cost

Answer 1

The answer is, Fertilizer A = 20, Fertilizer B = 56.

What is fertilizer and example?

• Among the organic fertilizers that occur naturally are manure, slurry, worm castings, peat, seaweed, and guano. Crops made of green manure are also raised to enrich the soil with nutrients.
• Organic fertilizers also include naturally occurring minerals including mine rock phosphate, sulfate of potash, and limestone.

Set up the equations:

• Nitrogen: 8x + 5y ≥ 440.
• Phosphorous: 2x + 5y ≥ 260.
• Potassium: 4x + 5y ≥ 360.

Find the vertices:

• It is easiest to graph the equations to find the vertices. (see attachment).
• You can also solve each system of equations to find the intersected points.

The following satisfy the "greater than or equal to" requirement:

• (0, 88)  = y-intercept of Nitrogen equation.
• (20, 56) = intersection of Nitrogen and Potassium equations.
• (50, 32)  = intersection of Phosphorous and Potassium.
• (130, 0)  = x-intercept of Potassium.

Use vertices in cost function C(x) to find the minimum:

• C(x) = $30x + $20y.
• (0, 88): $30(0) + $20(88) = $1760.
• (20, 56): $30(20) + $20(56) = $1720    = This is the minimum!
• (50, 32): $30(50) + $20(32) = $2140.
• (130, 0): $30(130) + $20(0) = $3900.

The minimum cost occurs when 20 bags of Fertilizer A and 56 bags of Fertilizer B are purchased.  

Learn more about Fertilizer here:

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