Question

A farmer uses two types of fertilizers. A 50-lb bag of Fertilizer A contains 8 lb of nitrogen, 2 lb of phosphorus, and 4 lb of potassium. A 50-lb bag of Fertilizer B contains 5 lb each of nitrogen, phosphorus, and potassium. The minimum requirements for a field are 440 lb of nitrogen, 260 lb of phosphorus, and 360 lb of potassium. If a 50-lb bag of Fertilizer A costs $30 and a 50-lb bag of Fertilizer B costs $20, find the amount of each type of fertilizer the farmer should use to minimize his cost while still meeting the minimum requirements.

Answer 1

Answer: Fertilizer A = 20, Fertilizer B = 56

Step-by-step explanation:

Step 1: Set up the equations

$\begin{array}{c|c|c|c}& FertilizerA&FertilizerB&Quantity Required\\Nitrogen&8&5&440\\Phosphorous&2&5&260\\Potassium&4&5&360\\\end{array}$

Nitrogen: 8x + 5y ≥ 440

Phosphorous: 2x + 5y ≥ 260

Potassium: 4x + 5y ≥ 360

Step 2: 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

Step 3: 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.