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? $
maximize
150 x + 50 y
x + y <=70
20 x + 60 y <= 3000
x>= 0
y>= 0
x+y >=1 to fill in input form only
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=1e692c6f72587b2cbd3e7be018fd8960&title=Linear%20Programming%20Calculator&theme=blue
result 10500 at (70,0)
which we could have seen without doing the problem :) The potato seed costs less and the profit is more
Thanks!
You are welcome.
To determine how many acres of potatoes and corn the farmer should plant, we can set up a system of equations. Let's use x to represent the number of acres of potatoes and y to represent the number of acres of corn.
The total cost of the seed is $20 per acre for potatoes and $60 per acre for corn, and the farmer has $3000 to spend. So, we can write the first equation as:
20x + 60y = 3000 (Equation 1)
The profit per acre for potatoes is $150 and for corn is $50. Therefore, the total profit can be calculated as:
Profit = 150x + 50y
To maximize the profit, we need to find the maximum value of the profit function. This requires identifying the values of x and y that satisfy both the profit function and the budget constraint.
To solve this, we can use the method of substitution or elimination. Here, we will use the method of substitution:
First, let's solve Equation 1 for x:
20x = 3000 - 60y
x = (3000 - 60y) / 20
x = 150 - 3y (Equation 2)
Now, substitute the value of x from Equation 2 into the profit equation:
Profit = 150(150 - 3y) + 50y
Profit = 22500 - 450y + 50y
Profit = 22500 - 400y (Equation 3)
Now we have a single-variable profit equation. To maximize this equation, we can take the derivative of Equation 3 with respect to y and set it equal to zero:
d(22500 - 400y)/dy = 0
-400 = 0
y = 56.25
Now, substitute the value of y back into Equation 2 to find x:
x = 150 - 3(56.25)
x = 150 - 168.75
x = -18.75
Since we cannot have negative acres, we discard the negative values:
Therefore, the farmer should plant 56 acres of corn and 18 acres of potatoes.
To find the maximum profit, substitute these values of x and y into the profit equation:
Profit = 22500 - 400(56.25)
Profit = 22500 - 22500
Profit = $0
Thus, the maximum profit is $0.