SOLUTION: A farmer has a field of 70 acres in which he plants potatoes and corn. The seed for potatoes costs $20/acre, the seed for corn costs $60/acre, and the farmer has set aside $3000 to

Algebra ->  Linear-equations -> SOLUTION: A farmer has a field of 70 acres in which he plants potatoes and corn. The seed for potatoes costs $20/acre, the seed for corn costs $60/acre, and the farmer has set aside $3000 to      Log On


   



Question 1140867: A farmer has a field of 70 acres in which he plants potatoes and corn. The seed for potatoes costs $20/acre, the seed for corn costs $60/acre, and the farmer has set aside $3000 to spend on seed. The profit per acre of potatoes is $150 and the profit for corn is $50 an acre. How many acres of each should the farmer plant?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
total acres are 70.
farmer has seeds to plan either potatoes or corn or both.
seed for potatoes is 20 dollars per acre.
see for corn is 60 dollars per acre.
total money for seed is 3000.
profit per acre of potatoes is 150 dollars.
profit per acre of corn is 50 dollars.

this problem can be solved graphically.

x = number of acres of potatoes
y = number of acres of corn.

objective function is profit = 150 * x + 50 * y

this is the function you want to maximize.

constraints are:

x >= 0
y >= 0
20 * x + 60 * y <= 3000
x + y <= 70

using the desmos.com calculator, you would graph the opposite of these inequalities.

the area on the graph that is not shaded is your region of feasibility.

the corner points of this area will be where the maximum profit lies.

the graph looks like this:

$$$

the corner points of the reasible region are:

(0,50), (30,40), (70,0), (0,0)

profit at each of these corner points are:

0 * 150 + 50 * 50 = 2500 for (0,50)
30 * 150 + 40 * 50 = 6500 for (30,40)
70 * 150 + 0 * 50 = 10500 for (70,0) *****
0 * 150 + 0 * 50 = 0 for (0,0)

maximum profit is attained when all 70 acres are planted by potatoes.

all constraints need to be satisfied as well.

constraints are:

x >= 0 = satisfied
y >= 0 = satisfied
20 * x + 60 * y <= 3000 = 20 * 70 + 60 * 0 <=3000 = 1400 <= 3000 = satisfied
x + y <= 70 = 70 + 0 <= 70 = 70 <= 70 = satisfied.

all constraints were satisfied, therefore the solution looks good.

the solution is that all 70 acres should be planted by potatoes for maximum profit, based on the information provided in the problem.



Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

1.   The seed for potatoes is potatoes itself.   There is NO other special seed for potatoes.

       (    It is for your general education.     )


2.   According to the condition,  "the seed for potatoes"  is cheaper for an acre,  while the profit per acre of potatoes is greater.

       So,  it is clear without any Math, that planting potatoes is the best strategy.