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 $80 and a 50-lb bag of Fertilizer B costs $30, find the amount of each type of fertilizer the farmer should use to minimize his cost C in dollars while still meeting the minimum requirements. (Let x represent the number of bags of Fertilizer A and y represent the number of bags of Fertilizer B.)

Answer 1

Answer:

The amount fertilizer A the farmer should use while still meeting the minimum requirement is approximately seventy-six 50-lb bags which would cost $6080.

The amount of fertilizer B the farmer should use while still meeting the minimum requirement is approximately seventy-one 50- lb bags which would cost $2120

Step-by-step explanation:

Fertilizer A

Let x represent the number of 50- lb bags needed for the minimum requirement

Cost of I 50-lb bag = $80

Cost of x 50-lb bag = $80x

1 50-lb bag contains 14lb nutrient ( 8lb Nitrogen+ 2lb Phosphorus + 4lb Potassium)

x 50-lb bag contains 1060lb nutrient (440lb N + 260lb P + 360lb K) which is the minimum requirement

x = 1060lb ÷ 14lb = 75.71

x = 76 ( to the nearest whole number)

76 50-lb bags would cost 76×$80 = $6080

Fertilizer B

Cost of a 50- lb bag is $30

Let y be the number of bags required

Cost of y bag is $30y

y = 1060÷ 15= 71

Cost = 71× $30 = $2130

Answer 2

Answer:

0 fertilizer A

88 fertilizer B

Step-by-step explanation:

Shown in the attachment