Question

Small farm has 70 acres of space to plant.farmer wants to grow corn and green beans. cost $60/acre to grow corn and $30 acre for green beans. Farmer has $1800 to use. Takes farmer 3 days to plant acre of corn and 4days to plant acre of green beans. 120 days available to plant vegetables. Farmer's profit is $180/acre of corn and $100/acre of green beans. How many acres should farmer allow for each vegetable to maximize profit?

Answer 1

Let x = no. of acres for corn
Let y = no. of acres for beans
:
Write an equation for each constraint, put each equation the slope intercept for for graphing
:
The Acerage equation
x + y =< 70
y = -x + 70, (Red)
:
Cost equation
60x + 30y =< 1800
30y = -60x + 1800
y = -2x + 60, Green
:
Time equation



3x + 4y =< 120
4y = -3x + 120
y = -.75x + 30 (blue)
:
Obviously, all values are positive
x => 0
y => 0
;
Graph all 3 equations on the same grid

:
From the graph we can see an intersection at x = 24, y = 12
Check and you can see these values will give us a cost of $1800 and time of 120 days
We can ignore the acreage equation, only 36 acres can be planted
:
"When the farmer sells the vegetables, he can expect to make a profit of $180 per acre for the corn and $100 per acre for the green beans."
Profit:
24(180) + 12(100) =
4320 + 1200 = $5520 max profit
:
But another corner is x=30, y=0; 30 acres of corn and no beans
It also will cost $1800 and but only 90 days to plant
Profit: 30(180) = $5400, slightly less, but less time required
:
You can check the 3rd corner. x = 0; y = 30, 30 acres of beans and no corn

Answer 2

I don't understand how to get constraints for this problem and the steps to solve

Answer 3

Ok what are the points on the graph and what does the graph look like