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Type 226f or Say 226f menu kitchen menu how many menu kitchen menu optimization
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Gaussian Elimination
To solve 3 - simultaneous linear equations
The health club serves a special meal consisting of three kinds of food called menu A, menu B, and menu C. Menu A consists of 20 grams of carbohydrates, 2 grams of fats and 4 grams of proteins. Menu B consists of 5 grams of carbohydrates, 1 grams of fat and 2 grams of proteins. Finally Menu c consists of 80 grams of carbohydrates, 3 grams of fat and 8 grams of proteins. If the total grams of carbohydrates is 140 grams , for fats the total grams is 11 gram and for protein the total grams is 24 grams. How many menu A, menu B, and menu C can the health clud can make?
X +
Y +
Z =
total grams for carbohydrates
Equation 1 or constraint number 1 that must be satisfied when you check for validation
X +
Y +
Z =
total grams for fats
Equation 2 constraint number 2 that must be satisfied when you check for validation
X +
Y +
Z =
total grams for proteins
Equation 3 constraint number 3 that must be satisfied when you check for validation
HOW TO USE: Simply enter new numbers.
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 with very small error.
Answer, menu A, x =
Answer, menu B, y =
Answer, menu C, z =
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 1.1
| (20/20) x | (5/20) y | (80/20 z | = | 140/20 | Equation 1 multiply by 1/20 |
| 1 x | 0.25 y | 4 z | = | 7 | Equation 1 x (1/20) = equation 1.1 |
Eliminating X ; meaning make it zero.
X +
Y +
Z =
Equation 2.2
Reason Operation 3. A multiple of one equation maybe added to another equation. Multiply equation 1 by - 40 then add to Equation 2
| 2 x | 1 y | 3 z | = | 11 | Equation 2 |
| - 2 x | - 0.5 y | - 8 z | = | - 14 | Equation 1.1 x (-2) = equation 1.1 |
| 0 x | 0.5 y | -5 z | = | -3 | Equation 1.1 added to Equation 2 |
| 0 x | 10 y | -100 z | = | -60 | Expand by multiplying with 20 = Equation 2.2 |
Eliminating X ; meaning make it zero.
X +
Y +
Z =
Equation 3.1
Reason Operation 3. Add equation 1.2 to Equation 3.1
| 4 x | 2 y | 8 z | = | 24 | Equation 3 |
| -4 x | -1 y | -16 z | = | -28 | Equation 1.1 x (-4) = equation 1.2 |
| 0 x | 1 y | -8 z | = | -4 | Equation 1.1 + Equation 3 = equation 1.2 |
| 0 x | 20 y | -160 z | = | -80 | Equation 1.2 multiply by 20 = 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 2.2
Reason Operation 2. Both sides of Equation 2.1 multiply by 1/20.
| 0 x | 10/10 * y | 100/10 * z | = | -60/10 | Equation 2.2 multiply by 1/10 |
| 0 x | 1 y | -10 z | = | -6 | 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 = 400 not 40 as shown by manual calculation below is because the algorithm did not simplify by dividing by 10. So if you divide by 10 equation 3.2 you will get the same answer shown below using manual calculation.
| 0 x | - 20 * y | - 20 * 10 z | = | -20 * -6 | Equation 2.2 x (-20) = Equation 2.3 |
| 0 x | -20 y | 200 z | = | 120 | Equation 2.3 |
| 0 x | 20 y | -160 z | = | -80 | Equation 3.1 |
| 0 x | + 0 y | + 40 z | = | 40 | 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 above, 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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