Question 1198953: 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?
Acres of potatoes=
Acres of corn=
What is the maximum profit $?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
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?
Acres of potatoes=
Acres of corn=
What is the maximum profit $?
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The seed of potatoes ($20/acre) is cheaper than the seed of corn ($60/acre).
The profit per acre of potatoes ($150) is greater than the profit for corn per acre ($50).
It makes it clear that the most aggressive strategy should be applied, which is
to plant the potatoes as much as possible.
There are enough money to buy the potatoes seed for 70 acres 70*20 = 1400 dollars.
So, the optimal strategy is to plant 70 acres of potatoes and do not plant the corn, at all.
ANSWER. Acres of potatoes = 70. Acres of corn = 0. Profit = 70*150 = 10500 dollars.
Solved.
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
x = number of acres of potatoes
y = number of acres of corn
These are real numbers such that x ≥ 0 and y ≥ 0
Since the farmer has at most 70 total acres to work with, we can say:
x+y ≤ 70
20x = cost of just the potatoes
60y = cost of just the corn
20x+60y = total cost ≤ 3000
20x+60y ≤ 3000
20(x+3y) ≤ 3000
x+3y ≤ 3000/20
x+3y ≤ 150
System of inequalities
Use a graphing tool (such as Desmos or GeoGebra), or do so by hand, to create a graph like this
The blue shaded region represents the set of (x,y) points that satisfy all of the inequalities mentioned in the system above.
Points on the boundary are included in the shaded solution set.
The corner points are:
A = (0, 50)
B = (30, 40)
C = (70, 0)
D = (0, 0)
Each corner can be found using algebraic methods to solve systems of equations.
For example, use algebra to solve the system
to determine the location of corner point B(30,40).
Once we have these corner points established, we plug each of them into the profit function
P(x,y) = 150x+50y
where,
150x = profit from just the potatoes only ($150 per acre)
50y = profit from just the corn only ($50 per acre)
You should find the following
| Point | Coordinates | Profit | | A | (0,50) | $2500 | | B | (30,40) | $6500 | | C | (70,0) | $10500 | | D | (0,0) | $0 |
Point C is the winner in terms of max profit.
This makes sense because the profit on potatoes is three times as much as corn.
Therefore, the farmer should use all 70 acres to plant nothing but potatoes.
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Answers:
Acres of potatoes = 70
Acres of corn = 0
Max Profit = $10,500
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