Question

A farmer owns a 100 acre farm and plans to plant at most three crops. The seed for crops A,B, and C costs $40, $20, and $30 per acre, respectively. A maximum of $3200 can be spent on seed. Crops A,B, and C require 1,2, and 1 workdays per acre, respectively, and there are maximum of 160 workdays available. If the farmer can make a profit of $100 per acre on crop A, $300 per acre on crop B, and $200 per acre on crop C, how many acres of each crop should be planted to maximize profit

Answer 1

Answer:

The number of acre per crop for maximum profit is;

Crop A = 0 acres

Crop B = 60 acres

Crop C = 40 acres

Profit, P = $26,000

Step-by-step explanation:

We plant A in X acres

B in Y acres and

C in Z acres

Therefore, X + Y + Z ≤ 100

we work A for 1·X workdays

B for 2·Y workdays and

C for 1·Z workdays

Where X + 2·Y + Z ≤ 160

$40·X for crop A

$20·Y for crop B and

$30·Z for crop C is spent whereby

$40·X + $20·Y + $30·Z ≤ $3200

The farmer makes

$100·X from crop A

$300·Y from crop B and

$200·Z from crop C

P = $100·X + $300·Y + $200·Z

Therefore, we have three equations with three unknowns solving the equations simultaneously, we have

X + Y + Z = 100....................................................(1)

X + 2·Y + Z = 160.................................................(2)

$40·X + $20·Y + $30·Z = $3200....................(3)

By subtracting equation (1) from (2) gives Y = 60 acres

Multiplying equation (1) by 40 and subtracting from (3) we  have Z = -40

and therefore, Y = 80

If he plants only B the farmer has 2 work day per acre and since there is a max of 160 days, he can only plant on 80 acres, therefore, total profit = $300 × 80 = $24,000

Comparing the profit per acre to the seed cost we have Profit for seed A = $100 while cost = $40 per acre,

If we remove seed A we have

Y + Z = 100....................................................(4)

2·Y + Z = 160.................................................(5)

$40·X + $20·Y + $30·Z = $3200....................(3)

Solving equation (4) and (5), we have Y = 60 acres and Z = 40 acres

Therefore, the profit becomes

$300 × 60 + $200 × 40 = $26,000.