A farmer with 1200 acre is considering planting 3 different crops corn, soybeans, and oats. The farmer wants to use all 1200 acres. seed corn costs $20 per acre, while soybeans and oats costs $50 and $12 respectively. The farmer has the $40000 available to buy seed and intends to spend it all. 1.Use the information above to formulate two linear equations with 3 unknowns and solve it. 2. write two reasonable solutions. 3.suppose in the fall when the crops mature, the farmer can bring in revenue of $100 per acre for corn, $300 per acre for soybean and $80 per acre for oats which of the two solutions given above would have resulted in a larger revenue.
Dealing with the same problem rn and I'm having trouble solving it
If the crops mature, does the farmer still have 40000 available to spend for seeds?
To solve this problem, we need to set up a system of linear equations based on the given information. Let's define the variables:
- Let x represent the number of acres planted with corn.
- Let y represent the number of acres planted with soybeans.
- Let z represent the number of acres planted with oats.
We know that the farmer wants to use all 1200 acres, so the first equation is:
x + y + z = 1200 (Equation 1)
The cost of seed corn is $20 per acre, soybeans cost $50 per acre, and oats cost $12 per acre. The farmer has $40000 available, and they intend to spend it all. Therefore, the second equation is:
20x + 50y + 12z = 40000 (Equation 2)
To solve this system of equations, we can use substitution or elimination method. Let's solve it using the substitution method.
From Equation 1, we can express x in terms of y and z:
x = 1200 - y - z
Substituting this value of x into Equation 2, we have:
20(1200 - y - z) + 50y + 12z = 40000
24000 - 20y - 20z + 50y + 12z = 40000
30y - 8z = 16000 (Equation 3)
Now we have a system of two linear equations with two variables:
Equation 1: x + y + z = 1200
Equation 3: 30y - 8z = 16000
To solve this system, we can use the elimination method by multiplying Equation 3 by 10 to make the coefficients of y in both equations equal:
10(30y - 8z) = 10(16000)
300y - 80z = 160000 (Equation 4)
Now we have a new system:
Equation 1: x + y + z = 1200
Equation 4: 300y - 80z = 160000
To eliminate z variable, we can multiply Equation 1 by 80:
80(x + y + z) = 80(1200)
80x + 80y + 80z = 96000 (Equation 5)
Now we have the following system:
Equation 5: 80x + 80y + 80z = 96000
Equation 4: 300y - 80z = 160000
To eliminate z variable, we can subtract Equation 4 from Equation 5:
80x + 80y + 80z - (300y - 80z) = 96000 - 160000
80x - 220y = -64000 (Equation 6)
We now have a new equation:
Equation 6: 80x - 220y = -64000
To solve for x and y, we can rearrange Equation 6:
80x = 220y - 64000
x = (220y - 64000) / 80
x = 2.75y - 800 (Equation 7)
Now we have an equation expressing x in terms of y.
To find two reasonable solutions, we can start by assigning values to y and calculating x and z using Equations 7, 1, and the constraint that x, y, and z all equal or greater than zero.
Solution 1:
Let y = 200 acres (arbitrary value)
Substituting y = 200 into Equation 7:
x = 2.75(200) - 800 = 350 acres
Using Equation 1 to find z:
z = 1200 - x - y = 1200 - 350 - 200 = 650 acres
Solution 2:
Let y = 400 acres (arbitrary value)
Substituting y = 400 into Equation 7:
x = 2.75(400) - 800 = 600 acres
Using Equation 1 to find z:
z = 1200 - x - y = 1200 - 600 - 400 = 200 acres
So, we have two solutions:
Solution 1: x = 350 acres, y = 200 acres, z = 650 acres
Solution 2: x = 600 acres, y = 400 acres, z = 200 acres
Now let's determine which solution would result in a larger revenue based on the fall crop prices.
Solution 1:
Revenue from corn: 100 * x = 100 * 350 = $35,000
Revenue from soybeans: 300 * y = 300 * 200 = $60,000
Revenue from oats: 80 * z = 80 * 650 = $52,000
Total revenue: $35,000 + $60,000 + $52,000 = $147,000
Solution 2:
Revenue from corn: 100 * x = 100 * 600 = $60,000
Revenue from soybeans: 300 * y = 300 * 400 = $120,000
Revenue from oats: 80 * z = 80 * 200 = $16,000
Total revenue: $60,000 + $120,000 + $16,000 = $196,000
Therefore, Solution 2 would result in a larger revenue.
Let the acreage be
x: corn
y: soy
z: oats
x+y+z = 1200
20x+50y+12z = 40000
with your solutions, then evaluate
revenue = 100x+300y+80z