A farmer is going to divide her 50 acre farm between two crops. Seed for crop A costs $50 per acre. Seed for crop B costs $25 per acre. The farmer can spend at most $2,250 on seed. If crop B brings in a profit of $90 per acre, and crop A brings in a profit of $220 per acre, how many acres of each crop should the farmer plant to maximize her profit? Write the objective function then use the feasible region shown in the graph below to maximize it.
Let x = the number of acres of crop A
Let y = the number of acres of crop B
Why did the scarecrow win an award? Because he was outstanding in his field! Now, let's get down to business.
To maximize the farmer's profit, we need to create an objective function. Let's call it P for profit:
P = 220x + 90y
Next, we have some constraints based on the information given. Firstly, the farmer has a total of 50 acres of land, so the sum of crop A and crop B cannot exceed that:
x + y ≤ 50
Secondly, the farmer has a budget limit of $2,250 for seed costs. Since crop A costs $50 per acre and crop B costs $25 per acre, we can express this constraint as:
50x + 25y ≤ 2,250
Now we can use the graph to determine the feasible region. Since I'm just a Clown Bot and not very skilled at drawing graphs, I'm afraid you'll have to use your imagination or actually draw it out. The feasible region on the graph will be the shaded area that satisfies both constraints.
Once you have the feasible region established, you'll want to find the corner points of the region. These corner points will represent different combinations of acreage for crop A and crop B that are within the constraints. Lastly, you can substitute these corner points into the objective function P = 220x + 90y to determine which combination maximizes profit.
I hope this makes you smile, even if it's just a little!
The objective function is the function that calculates the profit based on the number of acres of each crop. In this case, the objective function can be written as:
Profit = (Profit per acre of crop A * Number of acres of crop A) + (Profit per acre of crop B * Number of acres of crop B)
Using the given information:
Profit = (220x) + (90y)
Now, let's consider the constraints:
1. The total number of acres of both crops cannot exceed 50 acres:
x + y <= 50
2. The total cost of the seeds cannot exceed $2,250:
(50 * x) + (25 * y) <= 2,250
3. The number of acres of each crop cannot be negative:
x >= 0
y >= 0
Based on these constraints, we can plot the feasible region on a graph and find the maximum value of the objective function within that region.
To find the number of acres of each crop that the farmer should plant to maximize her profit, we can set up a system of equations based on the given information.
Let's start by defining the variables:
x = number of acres of crop A
y = number of acres of crop B
Now, let's set up the equations based on the given information:
1. The total number of acres planted: x + y = 50 (since the farmer is dividing her 50-acre farm between two crops)
2. The total cost of seed: 50x + 25y ≤ 2250 (since the farmer can spend at most $2,250 on seed)
3. The profit from crop A: 220x
4. The profit from crop B: 90y
The objective function, which represents the total profit, can be written as:
Total Profit = 220x + 90y
To graphically represent this problem, we can plot the feasible region on a graph. The feasible region represents the area where the constraints are satisfied.
On the graph, the x-axis represents the number of acres of crop A, and the y-axis represents the number of acres of crop B.
To graph the first constraint, x + y = 50, we can plot points on the graph where x + y equals 50. This will give us a straight line passing through points (0, 50) and (50, 0).
To graph the second constraint, 50x + 25y ≤ 2250, we can use the slope-intercept form of the equation: y ≤ (2250 - 50x) / 25. This will give us a shaded region below the line.
Once we have the feasible region graphed, we can find the coordinates of the vertices of the region and substitute them into the objective function to determine which combination of crop A and crop B yields the maximum profit.
By solving this optimization problem, the farmer can determine the number of acres of each crop to plant in order to maximize her profit.
cost: 50x+25y
profit: 220x + 90y
So, you want to
maximize p = 220x+90y subject to:
50x+25y <= 2250
0 <= x
0 <= y
x+y = 50
So, draw the region and evaluate p(x,y) on the vertices of the boundary.