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Farming Output Optimization
Automated Intelligent (AI) Formula
NO MORE LEARNING LOSS PROBLEM!
A farmer has 200 acres of land suitable for cultivating crop A, crop B, and crop C. The cost per acre of cultivating crops A, B, and C is $40, $60, and $80 respectively. The farmer has $12,600 available for cultivation. Each acre of crop A requires 20 labor-hours, each acre of crop B requires 25 labor-hours, and each acre of crop C requires 40 labor-hours. The farmer has a maximum of 5950 labor hours available. If she wishes to use all of her cultivatable land, the entire budget, and all the labor available, how many acres of each crop could she plant?
How to create the mathematical equations? Focus on constraint that needs to be satisfied.
Constraint number 1: the total number of acres to plant crop A, crop B and crop C is equal to 200 acres. First step assign variable letter x = number of acres needed to plant crop A
letter y = number of acres needed to plant crop B
letter z = number of acres needed to plant crop C
Written mathematically constraint number 1 as x + y + z = 200 acres
Constraint number 2: the total available budget to plant crop A, crop B, and crop C is $12,600
What is the budget to plant one acre of crop A ? answer is $40. Written mathematically 40x; What is the budget to plant one acre of crop B? answer is $60. Written mathematically 60y; What is the budget to plant one acre of crop C? answer is $80. Written mathematically 80z. Written mathematically constraint number 2 , you will arrived at this mathematical equation as 40x + 60y + 80 z = 12,600 budget
Constraint number 3: the total labor hours to plant crop A, crop B, and crop C is 5,950 hours. What is the labor hour needed to plant crop A in one acre land? answer 20 hours, written mathematically as 20x ; What is the labor hour needed to plant crop B in one acre land ? answer is 25 hours, written mathematically as 25y; What is the labor hour needed to plant crop C in one acre land? answer is 40 hours, written mathematically as 40z.
Written mathematically constraint number 3 you will get 20x + 25y + 40z = 5,950 hours.
X +
Y +
Z =
acres
Equation 1 or constraint 1 that needs to be satisfied when you do the final validation check.
X +
Y +
Z =
budget
Equation 2 or constraint 2 that needs to be satisfied when you do the final validation check.
X +
Y +
Z =
hours
Equation 3 or constraint 3 that needs to be satisfied when you do the final validation check.
These are solved using Cramer's Rule; Keep in mind that sometimes Cramer's Rule doesn't work. Therefore always substitute the value of x, y, and z to equation 1, 2, and 3 to check if they are valid answer.
To validate your answer substitute the values of x, y, and z to your original equation 1 , equation 2, and equation 3. The answer must satisfy all the equations or the three specified constraint with very small error.
Answer, crop A number of acres, X =
Answer, crop B number of acres, Y =
Answer, crop C number of acres, Z =
You can also solve this automatically ,
Using Wolfram Alpha to solve three simultaneous linear equations:
x + y + z = 200, 40 x + 60 y + 80 z = 12600, 20 x + 25 y + 40 z = 5950
The third way to solve this is using Gaussian Elimination Method
Result of Pivot 1, which means making column 1 follow the pattern [ 1, 0 , 0 ]
Result of Pivot 1, eliminating x and x in column 1.
X +
Y +
Z =
Equation 1
Eliminating X ; meaning make it zero.
X +
Y +
Z =
Equation 2.1
Reason Operation 3. A multiple of one equation maybe added to another equation. Multiply equation 1 by - 40 then add to Equation 2
| 40 x | 60 y | 80 z | = | 12,600 | Equation 2 |
| - 40 x | - 40 y | - 40 z | = | - 8,000 | Equation 1 x (-40) = equation 1.1 |
| 0 x | 20 y | 40 z | = | 4,600 | Equation 1.1 added to Equation 2 |
| 0 x | 1 y | 2 z | = | 230 | Simplify by dividing 20 = Equation 2.2 |
Eliminating X ; meaning make it zero.
X +
Y +
Z =
Equation 3 .
Reason Operation 3. Add equation 1.2 to Equation 3.1
| 20 x | 25 y | 40 z | = | 5,950 | Equation 3 |
| -20 x | -20 y | -20 z | = | -4,000 | Equation 1 x (-20) = equation 1.2 |
| 0 x | 5 y | 20 z | = | 1,950 | Equation 1.2 added to Equation 3 = Equation 3.1 |
Result of Pivot 2, which means making column 2 follow the pattern [ 1, 1 , 0 ]
Result of Pivot 2 eliminating in equation 3 above.
Then converting in equation 2.1 to one (1) in column 2.
X + Y + Z = Equation 1
X +
Y +
Z =
Equation 2.1
Reason Operation 2. Both sides of Equation 2.1 multiply by 1/20.
| 0 x | 20/20 * y | 40/20 * z | = | 4,600/20 | Equation 2.1 multiply by 1/20 |
| 0 x | 1 y | 2 z | = | 230 | Equation 2.2 |
X +
Y +
Z =
Equation 3.2
Reason Operation 3. Multiply Equation 2.2 by -5 and add to equation 3.1
The reason why Z = 200 not 10 as shown by manual calculation below is because the algorithm did not simplify by dividing by 20. So if you divide by 20 equation 3.2 you will get the same answer shown below using manual calculation.
| 0 x | - 5 * y | - 5 * 2 z | = | -5 * 230 | Equation 2.2 x (-5) = Equation 2.3 |
| 0 x | -5 y | -10 z | = | -1,150 | Equation 2.3 |
| 0 x | 5 y | 20 z | = | 1,950 | Equation 3.1 |
| 0 x | + 0 y | + 10 z | = | 800 | Equation 2.3 + Eq. 3.1 = Equation 3.2 |
ANSWER: Z = .
Using back substitution , substitute Z = in equation 2.1 above, you will get
Y = .
Finally substitute Z = , Y = in equation 1 below, you will get X = .
Important definition to remember about row echelon. The process of reducing augmented matrix in row echelon form is called Gaussian elimination
1. The first nonzero entry in each row is always 1
2. A row with more leading zero entries compare to the previous row must be located below.
3. A row with all zero entries must be below the rows having nonzero entries.
Gaussian elimination will not work properly if one of the definition is violated.
Operation 1 - The order in which any two equations are written may be interchanged.
Operation 2 - Both sides of the equation may be multiplied by the same nonzero real number.
Operation 3 - A multiple of one equation may be added to another equation.
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