To determine To calculate: The number of 100-gram units of each ingredient to be used in stir fry and convert to the kitchen with the help of the table. Explanation Given: A stir fry which contains more than 50 grams of protein, less than 10 grams of carbohydrates and more than 3.5 grams of vitamin C and at least 1 tablespoon which is 4.5 grams of oil is used in the recipe. Conversion of ingredients of various food that is listed in the tabular form: Food Serving size Beef 6 oz = 170 g Egg 1 egg = 61 g Feta Cheese 1 4 cup = 38 g Lettuce 1 2 cup = 28 g Milk 1 cup = 244 g Oil 1 Tbsp = 13.5 g Onion 1 Onion = 110 g Salad Dressing 1 cup = 250 g Soy Sauce 1 Tbsp = 18 g spinach 1 cup = 180 g Tomato 1 tomato = 123 g The amounts of calorie and nutrients are per 100 gram of the ingredients are listed in the form of a table: Nutrient (units) Beef Oil Onion Soy sauce Calories (kcal) 215 884 38 60 Protein (g) 26 0 1.16 10.51 Carbohydrates (g) 0 1 8.63 5.57 Vitamin C (g) 0 0 6.4 0 Formula used: Solution: Consider minimization problems and formulate into a system of linear equations Take y 1 be the number of 100-gram units of beef, then take y 2 be the number of 100-gram units of oil, y 3 be the number of 100-gram units of onion and take y 4 be the number of 100-gram units of soy sauce. Consider the amounts of calorie and nutrients is per 100 gram for the ingredients are listed in the form of a table: Nutrient (units) Beef Oil Onion Soy sauce Calories (kcal) 215 884 38 60 Protein (g) 26 0 1.16 10.51 Carbohydrates (g) 0 1 8.63 5.57 Vitamin C (g) 0 0 6.4 0 Take, w be the objective function to minimize the calories in stir fry which is given 215 kilocalories of beef, 884 kilocalories of oil, 38 kilocalories of onion, and 60 kilocalories of soy sauce so the objective function is: w = 215 y 1 + 884 y 2 + 38 y 3 + 60 y 4 Stir-fry contains less than 10 gram of carbohydrates, so the constraints for calories from the above table is: y 2 + 8.63 y 3 + 5.57 y 4 < 10 Stir-fry contains more than 50 gram of protein, so the constraints for calories from the above table is: 26 y 1 + 1.16 y 3 + 10.51 y 4 > 50 Stir-fry contains more than 3.5 gram of vitamin C, so the constraints for calories from the above table is: 6.4 y 3 > 3.5 Oil is at least 1 tablespoon which is 4.5 grams of oil from the table it converts into gram As it is considered the number of 100-gram units therefore for 4.5 grams, 100 gram = 1 unit 1 gram = 1 100 4.5 gram = 1 100 × 4.5 = .045 So the constraints for oil is: y 2 ≥ .045 Therefore, formulate the problem which is: Minimize, w = 215 y 1 + 884 y 2 + 38 y 3 + 60 y 4 Subject to, y 2 + 8.63 y 3 + 5.57 y 4 < 10 , 26 y 1 + 1.16 y 3 + 10.51 y 4 > 50 , 6.4 y 3 > 3.5 , y 2 ≥ .045 And, y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 , y 4 ≥ 0 Use two stages method there are various steps which have to follow: Step (1): modify each equation constraint by an equivalent pair of inequality constraint Step (2): write each constraint with a positive constant. Step (3): if minimization problem then converts into maximization by replace z = − w . Step (4): add or subtract surplus variables according to the need of the constraints into equations. Step (5): write initial simplex table. Step (6): find a basic feasible solution if a solution exists which is stage I Step (7): if basic feasible solution found then do the simplex method further which is a stage and get the optimal solution. Further, to convert linear inequality into the linear equation then take a slack variable and surplus variable which is done by adding and subtracting a non-negative variable to each constraint. If linear equality has less than or equal to sign then use slack variable by add as the equation is less than or equal to the value if it is equal then the slack variable is zero. If linear equality has greater than or equal to sign then use surplus variable by subtracting as the equation is greater than or equal to the value, if it is equal then the surplus variable is zero which is: y 2 + 8.63 y 3 + 5.57 y 4 + s 1 = 10 , 26 y 1 + 1.16 y 3 + 10.51 y 4 − s 2 = 50 , 6.4 y 3 − s 3 = 3.5 , y 2 − s 4 = .045 Where s 1 , s 2 , s 3 is greater than zero and s 4 ≥ 0 Then rewrite constraints by adding or subtracting the non-negative slack or surplus variables which are: Minimize, w = 215 y 1 + 884 y 2 + 38 y 3 + 60 y 4 Subject to, y 2 + 8.63 y 3 + 5.57 y 4 + s 1 = 10 , 26 y 1 + 1.16 y 3 + 10.51 y 4 − s 2 = 50 , 6.4 y 3 − s 3 = 3.5 , y 2 − s 4 = .045 And, y 1 ≥ 0 , y 2 ≥ 0 , s 1 > 0 , s 2 > 0 , s 3 > 0 s 4 ≥ 0 Convert minimization into maximization which is: z = − w So, Maximize, z = − 215 y 1 − 884 y 2 − 38 y 3 − 60 y 4 Subject to, y 2 + 8.63 y 3 + 5.57 y 4 + s 1 = 10 , 26 y 1 + 1.16 y 3 + 10.51 y 4 − s 2 = 50 , 6.4 y 3 − s 3 = 3.5 , y 2 − s 4 = .045 And, y 1 ≥ 0 , y 2 ≥ 0 , s 1 > 0 , s 2 > 0 , s 3 > 0 s 4 ≥ 0 Then convert the linear programming into a system of linear equations in which all the coefficient of variables are written on the left of the equal sign and the constants are on the right then convert the objective function into linear equations which are: 215 y 1 + 884 y 2 + 38 y 3 + 60 y 4 + z = 0 Then write the constraint and objective function into the augmented matrix which is in the form of a simplex table y 1 y 2 y 3 y 4 s 1 s 2 s 3 s 4 [ 0 1 8.63 5.57 1 0 0 0 10 26 0 1.16 10.51 0 − 1 0 0 50 0 0 6.4 0 0 0 − 1 0 3.5 0 1 0 0 0 0 0 − 1 .045 215 884 38 60 0 0 0 0 0 ] Here, the last row shows the coefficients of the objective function and last columns shows the constants. The basic variables correspond to columns which have an entry of 1 and rest is zero its value corresponds to the last column which is constant and the non-basic variables correspond to the other columns and its value is zero. In the above nonbasic variables are y 1 , y 2 , y 3 , y 4 and the basic variables are s 1 , s 2 , s 3 , s 4 . Here, y 1 = 0 , y 2 = 0 , y 3 = 0 , y 4 = 0 , s 1 = 10 , s 2 = − 50 , s 3 = − 3.5 , s 4 = − .045 Above simplex, the table has not a basic feasible solution as the basic variable has the negative value so first, convert basic variables into non negative then it is said to be basic feasible solution and stage I is completed. In the last row, all the entries except the last and second last entries are called indicators. The simplex table represents a system of five linear equations in nine variables which is y 1 , y 2 , y 3 , y 4 , s 1 , s 2 , s 3 , s 4 , z as there are more variable than equations then the system is dependent and it has infinitely many solutions. In the infeasible solution as s 2 has negative value, the only non-zero entry in its column is − 1 in the second row select any positive entry except the entry which is far from the right that is 50 . So, take y 1 column as pivot column. Now all the positive entry of pivot column is divided by the last column which is constant which is: First, 10 ÷ 0 = ∞ , Second, 50 ÷ 26 = 1.92 , Third, 3.5 ÷ 0 = ∞ , Fourth, .045 ÷ 0 = ∞ , After the row with the smallest quotient is called the pivot row which is 1.92 and it occurs in the second row. The entry in the pivot row and pivot column is called Pivot. y 1 y 2 y 3 y 4 s 1 s 2 s 3 s 4 [ 0 1 8.63 5.57 1 0 0 0 10 26 0 1.16 10.51 0 − 1 0 0 50 0 0 6.4 0 0 0 − 1 0 3.5 0 1 0 0 0 0 0 − 1 .045 215 884 38 60 0 0 0 0 0 ] In this, the pivot is 26 the pivot is selected then do the row operations in the initial simplex table to get the new simplex table by eliminate pivot column all value to zero except the pivot which is change by row transform into 1 Replace the second row by 1 26 times itself which is: R 2 → 1 26 R 2 Then, y 1 y 2 y 3 y 4 s 1 s 2 s 3 s 4 [ 0 1 8.63 5.57 1 0 0 0 10 1 0 .044 .404 0 − .038 0 0 1.92 0 0 6.4 0 0 0 − 1 0 3.5 0 1 0 0 0 0 0 − 1 .045 215 884 38 60 0 0 0 0 0 ] Replace the fifth row by sum itself and − 215 times of the second row which is: R 5 → R 5 − 215 R 2 Then, y 1 y 2 y 3 y 4 s 1 s 2 s 3 s 4 [ 0 1 8.63 5.57 1 0 0 0 10 1 0 .044 .404 0 − .038 0 0 1.92 0 0 6.4 0 0 0 − 1 0 3.5 0 1 0 0 0 0 0 − 1 .045 0 884 28.45 − 26.86 0 8.17 0 0 − 412.8 ] Repeat the process as the basic feasible solution does not exist which is: y 1 = 1.92 , y 2 = 0 , y 3 = 0 , y 4 = 0 , s 1 = 10 , s 2 = 0 , s 3 = − 3.5 , s 4 = − .045 In the infeasible solution as s 3 has negative value, the only non-zero entry in its column is − 1 in the third row select any positive entry except the entry which is far from the right that is 3.5 . So, take y 3 column as pivot column. Now all the positive entry of pivot column is divided by the last column which is constant which is: First, 10 ÷ 8.63 = 1.16 , Second, 1.92 ÷ .044 = 43.1 , Third, 3.5 ÷ 6.4 = .55 , Fourth, .045 ÷ 0 = ∞ . After the row with the smallest quotient is called the pivot row which is .55 , and it occurs in the third row. The entry in the pivot row and pivot column is called Pivot. y 1 y 2 y 3 y 4 s 1 s 2 s 3 s 4 [ 0 1 8.63 5.57 1 0 0 0 10 1 0 .044 .404 0 − .038 0 0 1.92 0 0 6.4 0 0 0 − 1 0 3

Mathematics with Applications and Mylab Math with Pearson EText -- Title-Specific Access Card Package

12th Edition

ISBN: 9780134862668

Author: Lial, Margaret L.

Publisher: Pearson Education Canada

See similar textbooks

Concept explainers

Inverse Normal Distribution

The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of tw…

Mean, Median, Mode

It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency o…

Z-Scores

A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow compar…

Videos

Question

Chapter 7, Problem 2CE

To determine

To calculate: The number of 100-gram units of each ingredient to be used in stir fry and convert to the kitchen with the help of the table.

Expert Solution & Answer

Chapter 7 Solutions

Mathematics with Applications and Mylab Math with Pearson EText -- Title-Specific Access Card Package

Ch. 7.1 - Checkpoint 1 Graph the given...Ch. 7.1 - Prob. 2CP Ch. 7.1 - Prob. 3CP Ch. 7.1 - Prob. 4CP Ch. 7.1 - Prob. 5CP Ch. 7.1 - Checkpoint 6 Graph the feasible region of the...Ch. 7.1 - Prob. 7CP Ch. 7.1 - Match the inequality with its graph, which is one...Ch. 7.1 - Match the inequality with its graph, which is one...Ch. 7.1 - Match the inequality with its graph, which is one...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.