Mathematics with Applications In the Management, Natural and Social Sciences (11th Edition)
Mathematics with Applications In the Management, Natural and Social Sciences (11th Edition)
11th Edition
ISBN: 9780321931078
Author: Margaret L. Lial, Thomas W. Hungerford, John P. Holcomb, Bernadette Mullins
Publisher: PEARSON
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Chapter 7, Problem 1CE
To determine

To calculate: The number of 100-gram units of each ingredient in a Greek salad and convert to the kitchen with the help of the table.

Expert Solution & Answer
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Answer to Problem 1CE

Solution:

About 12.41 gram of carbohydrates when the number of 100-gram units of feta cheese is .243, the number of 100-gram units of lettuce is 2.375, the number of 100-gram units of salad dressing is .3125 and the number of 100-gram units of tomato is 1.087.

Conversion of ingredients into the kitchen is:

16_ Cup of feta cheese, 414_ cup of lettuce, 18_ cup of salad dressing, 78_ of 1 tomato.

Explanation of Solution

Given:

The salad should less than 260 calories, and it weighs less than 400 grams and further dressed with at least 2 tablespoons. Calcium should over 210 milligrams, protein should over 6 grams.

Conversion of ingredients of various food that is listed in the tabular form:

Food

Serving size

Beef

6oz=170g

Egg

1egg=61g

Feta Cheese

14cup=38g

Lettuce

12cup=28g

Milk

1cup=244g

Oil

1Tbsp=13.5g

Onion

1Onion=110g

Salad Dressing

1cup=250g

Soy Sauce

1Tbsp=18g

spinach

1cup=180g

Tomato

1tomato=123g

The amounts of calorie and nutrients are per 100 grams for the ingredients are listed in the form of a table:

Nutrient (units)

Feta cheese

Lettuce

Salad Dressing

Tomato

Calories (kcal)

263

14

448.8

21

Calcium (g)

492.5

36

0

5

Protein (g)

10.33

1.62

0

.85

Carbohydrates (g)

4.09

2.37

2.5

4.64

The constants in the weight constraint are 4, not 400 grams.

Formula used:

Calculation:

Consider maximization problems and formulate into system of linear equations.

Take x1 be the number of 100-gram units of feta cheese, take x2 be the number of 100-gram units of lettuce, and x3 be the number of 100-gram units of salad dressing and take x4 be the number of 100-gram units of tomato.

Consider the amounts of calorie and nutrients is per 100 grams for the ingredients are listed in the form of a table:

Nutrient (units)

Feta cheese

Lettuce

Salad Dressing

Tomato

Calories (kcal)

263

14

448.8

21

Calcium (g)

492.5

36

0

5

Protein (g)

10.33

1.62

0

.85

Carbohydrates (g)

4.09

2.37

2.5

4.64

Take, z be the objective function to maximize the carbohydrate which is given 4.09-gram feta cheese, 2.37-gram lettuce, 2.5-gram salad dressing, and 4.64-gram tomato,

So, the objective function is:

z=4.09x1+2.37x2+2.5x3+4.64x4

Salad should less than 260 calories, so the constraints for calories from the above table is:

263x1+14x2+448.8x3+21x4<260

Calcium should over 210 milligrams, so the constraints for calcium from the above table is:

492.5x1+36x2+0x3+5x4>210

Protein should over 6 grams, so the constraints for protein from the above table is:

10.33x1+1.62x2+0x3+.85x4>6

The constants in the weight constraint are 4 so the constraints,

x1+x2+x3+x4<4

Consider the conversion table,

Conversion of ingredients of various food that is listed in the tabular form:

Food

Serving size

Beef

6oz=170g

Egg

1egg=61g

Feta Cheese

14cup=38g

Lettuce

12cup=28g

Milk

1cup=244g

Oil

1Tbsp=13.5g

Onion

1Onion=110g

Salad Dressing

1cup=250g

Soy Sauce

1Tbsp=18g

spinach

1cup=180g

Tomato

1tomato=123g

Salad dressing is at least 2 tablespoons which is 18 cup from the table it converts into gram which is:

1cup=250g18cup=250831.25g

As it is considered the number of 100-gram units therefore for 31.25 grams,

100gram=1unit1gram=1100

And,

31.25gram=1100×31.25=.3125

So, the constraints for salad dressing is:

x3.3125

Therefore, formulate the problem which is:

Maximize,

z=4.09x1+2.37x2+2.5x3+4.64x4

Subject to,

263x1+14x2+448.8x3+21x4<260492.5x1+36x2+0x3+5x4>21010.33x1+1.62x2+0x3+.85x4>6x1+x2+x3+x4<4

And,

x3.3125x10,x20,x30, and x40

Use two stages method to solve the maximization problem, then 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 add and subtract 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 subtract as the equation is greater than or equal to the value, if it is equal then the surplus variable is zero which is:

263x1+14x2+448.8x3+21x4+s1=260492.5x1+36x2+0x3+5x4s2=21010.33x1+1.62x2+0x3+.85x4s3=6x1+x2+x3+x4+s4=4

And,

x3+s5=.3125,x10,x20,x30, and x40

Where s1,s2,s3,s4 are greater to zero and s50

Then rewrite constraints by add or subtract the non-negative slack and surplus variables which are:

Maximize,

z=4.09x1+2.37x2+2.5x3+4.64x4

Subject to,

263x1+14x2+448.8x3+21x4+s1=260492.5x1+36x2+0x3+5x4s2=21010.33x1+1.62x2+0x3+.85x4s3=6x1+x2+x3+x4+s4=4

And,

x3+s5=0.3125x10,x20,x30,x40s1>0,s2>0,s3>0,s4>0,s50

Then, convert the linear programming into a system of linear equations in which all the coefficients 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:

4.09x12.37x22.5x34.64x4+z=0

Then, write the constraint and objective function into the augmented matrix which is in the form of a simplex table.

x1x2x3x4s1s2s3s4s5[26314448.82110000260492.536050100021010.331.620.850010061111000104001000001.31254.092.372.54.64000000]

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 x1,x2,x3,x4 and the basic variables are s1,s2,s3,s4,s5.

Here,

x1=0,x2=0,x3=0,x4=0, and

s1=260,s2=210,s3=6,s4=4,s5=.3125

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 six linear equations in ten variables which is:

x1,x2,x3,x4,s1,s2,s3,s4,s5,z.

As there are more variable than equations, then the system is dependent and it has infinitely many solutions.

In the infeasible solution as s2 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 210.

So, take x1 column as pivot column.

Now all the positive entry of pivot column is divided by the last column which is constant which is:

First, 260÷263=.98.

Second, 210÷492.5=.426.

Third, 6÷10.33=.58.

Fourth, 4÷1=4.

Fifth, .3125÷0=.

After the row with the smallest quotient is called the pivot row which is .426 and it occurs in the second row.

The entry in the pivot row and pivot column is called Pivot.

x1x2x3x4s1s2s3s4s5[26314448.82110000260492.536050100021010.331.620.850010061111000104001000001.31254.092.372.54.64000000]

In this, the pivot is 492.5 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 1492.5 times itself which is:

R21492.5R2

Then,

x1x2x3x4s1s2s3s4s5[26314448.821100002601.0730.0100.002000.42610.331.620.850010061111000104001000001.31254.092.372.54.64000000]

Replace the first row by sum itself and 263 times of the second row which is:

R1R1263R2

Replace the third row by sum itself and 10.33 times of the second row which is:

R3R310.33R2

Replace the fourth row by sum itself and 1 times of the second row which is:

R4R4R2

Replace the sixth row by sum itself and 4.09 times of the second row which is:

R6R6+4.09R2

Then,

x1x2x3x4s1s2s3s4s5[05.19448.818.371.526000147.961.0730.0100.002000.4260.8650.7460.0201001.590.9271.990.0020103.57001000001.312502.072.54.590.0080001.742]

Repeat the process as the basic feasible solution does not exist which is:

x1=.426,x2=0,x3=0, andx4=0

s1=147.96,s2=0,s3=1.59,s4=3.57, and s5=.3125

In the infeasible solution as s5 has negative value, the only non-zero entry in its column is 1 in the fifth row select any positive entry except the entry which is far from the right that is .3125.

So, take x3 column as pivot column.

Now all the positive entry of pivot column is divided by the last column which is constant which is:

First, 147.96÷448.8=.33,

Second, .426÷0=,

Third, 1.59÷0=,

Fourth, 3.57÷1=3.57, and

Fifth, .3125÷1=.3125.

After the row with the smallest quotient is called the pivot row which is .3125 and it occurs in the fifth row.

The entry in the pivot row and pivot column is called Pivot.

x1x2x3x4s1s2s3s4s5[05.19448.818.371.526000147.961.0730.0100.002000.4260.8650.7460.0201001.590.9271.990.0020103.57001000001.312502.072.54.590.0080001.742]

In this, the pivot is 1 the pivot is selected then do the row operations in the initial simplex table to get the new simplex table by elimination of pivot column all value to zero except the pivot which is change by row transform into 1

Replace the first row by sum itself and 448.8 times of the fifth row which is:

R1R1448.8R5

Replace the fourth row by sum itself and 1 times of the fifth row which is:

R4R4R5

Replace the sixth row by sum itself and 2.5 times of the fifth row which is:

R6R6+2.5R5

Then,

x1x2x3x4s1s2s3s4s5[05.19018.371.52600448.87.601.0730.0100.002000.4260.8650.7460.0201001.590.9270.990.0020113.26001000001.312502.0704.590.008002.52.52]

Repeat the process as the basic feasible solution does not exist which is:

x1=.426,x2=0,x3=.3125,x4=0,

And, s1=7.60,s2=0,s3=1.59,s4=3.26,s5=0.

In the infeasible solution, s3 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 1.59.

So, take x2 column as pivot column.

Now all the positive entry of pivot column is divided by the last column which is constant which is:

First, one is negative so not considered

Second, .426÷.073=5.83,

Third, 1.59÷.865=1.84,

Fourth, 3.26÷.927=3.52,

Fifth, .3125÷0=.

After the row with the smallest quotient is called the pivot row which is 1.84 and it occurs in the third row.

The entry in the pivot row and pivot column is called Pivot.

x1x2x3x4s1s2s3s4s5[05.19018.371.52600448.87.601.0730.0100.002000.4260.8650.7460.0201001.590.9270.990.0020113.26001000001.312502.0704.590.008002.52.52]

In this, the pivot is .865; 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 third row by 1.865 times itself which is:

R31.865R3

Then,

x1x2x3x4s1s2s3s4s5[05.19018.371.52600448.87.601.0730.0100.002000.426010.8610.0241.156001.8440.9270.990.0020113.26001000001.312502.0704.590.008002.52.52]

Replace the first row by sum itself and 5.19 times of the third row which is:

R1R1+5.19R3

Replace the second row by sum itself and .073 times of the third row which is:

R2R2.073R3

Replace the fourth row by sum itself and .927 times of the third row which is:

R4R4.927R3

Replace the sixth row by sum itself and 2.07 times of the third row which is:

R6R6+2.07R3

Then,

x1x2x3x4s1s2s3s4s5[00022.831.6606.040448.817.244100.05280.0038.084500.2916010.8610.0241.156001.844000.1910.0201.071111.551001000001.31250002.800.04162.3902.56.337]

Here,

x1=.2916,x2=1.844,x3=.3125,x4=0,

And, ,s1=17.244,s2=0,s3=0,s4=1.551,s5=0.

As all the basic variables are non-negative, so stage I is completed and then apply the simplex method for stage II.

In the simplex, choose the pivot entry as the pivot is the most negative indicator which is 2.80 as it occurs in the fourth column then column containing the most negative indicator is called pivot column which is x4.

Now all the positive entry of pivot column is divided by the last column which is constant which is:

First, 17.244÷22.83=.76,

Second, one is negative so not considered,

Third, 1.844÷.8615=2.14,

Fourth, 1.551÷.1913=8.11,

And, Fifth, .3125÷0=.

After the row with the smallest quotient is called the pivot row which is .76 as it occurs in the first row.

The entry in the pivot row and pivot column is called pivot

x1x2x3x4s1s2s3s4s5[00022.831.6606.040448.817.244100.05280.0038.084500.2916010.8610.0241.156001.844000.1910.0201.071111.551001000001.31250002.800.04162.3902.56.337]

In this, the pivot is 22.83; 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 first row by 122.83 times of itself which is:

R1122.83R1

Then,

x1x2x3x4s1s2s3s4s5[0001.043.0289.2646019.65.7553100.05280.0038.084500.2916010.8610.0241.156001.844000.1910.0201.071111.551001000001.31250002.800.04162.3902.56.337]

Replace the second row by sum itself and .0528 times of the first row which is:

R2R2+.0528R1

Replace the third row by sum itself and .861 times of the first row which is:

R3R3.861R1

Replace the fourth row by sum itself and .191 times of the first row which is:

R4R4.191R1

Replace the sixth row by sum itself and 2.80 times of the first row which is:

R6R6+2.80R1

Then,

x1x2x3x4s1s2s3s4s5[0001.043.0289.2646019.65.75531000.0023.0023.070501.038.33150100.0377.0007.9283016.9351.1930000.0084.0261.12212.7601.406001000001.31250000.1204.12253.12052.528.451]

In the simplex, choose the pivot entry as the pivot is the most negative indicator which is 3.12 as it occurs in the seventh column then column containing the most negative indicator is called pivot column which is s3.

Now all the positive entry of pivot column is divided by the last column which is constant which is:

First, one is negative so it is not considered,

Second, .3315÷.0705=4.7,

Third, one is negative so it is not considered,

Fourth, 1.406÷1.122=1.25,

And, Fifth, .3125÷0=.

After the row with the smallest quotient is called the pivot row which is 1.25 as it occurs in the fourth row.

The entry in the pivot row and pivot column is called pivot,

x1x2x3x4s1s2s3s4s5[0001.043.0289.2646019.65.75531000.0023.0023.070501.038.33150100.0377.0007.9283016.9351.1930000.0084.0261.12212.7601.406001000001.31250000.1204.12253.12052.528.451]

In this, the pivot is 1.122; 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 fourth row by 11.122 times of itself which is:

R411.122R4

Then,

x1x2x3x4s1s2s3s4s5[0001.043.0289.2646019.65.75531000.0023.0023.070501.038.33150100.0377.0007.9283016.9351.1930000.0075.02321.8912.461.253001000001.31250000.1204.12253.12052.528.451]

Replace the first row by sum itself and .2646 times of the fourth row which is:

R1R1+.2646R4

Replace the second row by sum itself and .0705 times of the fourth row which is:

R2R2.0705R4

Replace the third row by sum itself and .9283 times of the first row which is:

R3R3+.9283R4

Replace the sixth row by sum itself and 3.12 times of the first row which is:

R6R6+3.12R4

Then,

x1x2x3x4s1s2s3s4s5[0001.0418.02280.235719.0061.0871000.0028.00060.06291.2119.2430100.0447.02220.827119.21872.3750000.0075.02321.8912.461.253001000001.31250000.0998.050702.79745.09912.41]

From the above table, there is no negative indicator so the optimal solution exists and stage II is completed.

Here, the maximum value z appears in the lower right-hand corner which is 12.41 and the value of variables is:

x1=.243,x2=2.375,x3=.3125,x4=1.087,s1=17.244,s2=0,s3=0,s4=1.551,s5=0

So, the solution of the maximum value is:

z=4.09x1+2.37x2+2.5x3+4.64x4=4.09(.243)+2.37(2.37)+2.5(.3125)+4.64(1.087)=.9938+5.569+.78125+5.04=12.41

For, 12.41 gram of carbohydrates,

The number of 100-gram units of feta cheese is .243.

The number of 100-gram units of lettuce is 2.375.

The number of 100-gram units of salad dressing is .3125, and

The number of 100-gram units of tomato is 1.087.

Now, consider the conversion ingredients of various food which is:

Food

Serving size

Feta Cheese

14cup=38g

Lettuce

12cup=28g

Salad Dressing

1cup=250g

Tomato

1tomato=123g

As, .243 units of feta cheese is used for 100-gram units that is 24.3 gram then from the table,

38g=14cup1=14×3824.3=14×38×24.3=16cup

As 2.375 units of lettuce are used for 100-gram units that is 237.5 gram then from the table,

28g=12cup1=12×28237.5=12×28×237.5=174cup414

As, .3125 units of salad dressing is used for 100-gram units that is 31.25 gram then from the table,

250g=1cup1=125031.25=1250×31.25=18cup

As, 1.087 units of tomato is used for 100-gram units that is 108.7 gram then from the table,

123g=1tomato1=1123108.7=1123×108.7=78of 1tomato

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Chapter 7 Solutions

Mathematics with Applications In the Management, Natural and Social Sciences (11th Edition)

Ch. 7.1 - Prob. 4ECh. 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 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Prob. 9ECh. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Prob. 11ECh. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Prob. 17ECh. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Prob. 19ECh. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Prob. 22ECh. 7.1 - Prob. 23ECh. 7.1 - Graph each of the given linear inequalities. (See...Ch. 7.1 - Prob. 25ECh. 7.1 - Prob. 26ECh. 7.1 - Prob. 27ECh. 7.1 - Prob. 28ECh. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Prob. 32ECh. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Prob. 34ECh. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Prob. 37ECh. 7.1 - Prob. 38ECh. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Prob. 40ECh. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Prob. 43ECh. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Graph the feasible region for the given systems of...Ch. 7.1 - Prob. 47ECh. 7.1 - Prob. 48ECh. 7.1 - Prob. 49ECh. 7.1 - Prob. 50ECh. 7.1 - Prob. 51ECh. 7.1 - 52. Business A manufacturer of electric shavers...Ch. 7.1 - Prob. 53ECh. 7.1 - Prob. 54ECh. 7.1 - Prob. 55ECh. 7.1 - Prob. 56ECh. 7.2 - Checkpoint 1 Suppose the objective function in...Ch. 7.2 - Prob. 2CPCh. 7.2 - Prob. 3CPCh. 7.2 - Checkpoint 4 Use the region of feasible solutions...Ch. 7.2 - Prob. 5CPCh. 7.2 - Prob. 6CPCh. 7.2 - Exercises 1–6 show regions of feasible solutions....Ch. 7.2 - Exercises 1–6 show regions of feasible solutions....Ch. 7.2 - Exercises 1–6 show regions of feasible solutions....Ch. 7.2 - Exercises 1–6 show regions of feasible solutions....Ch. 7.2 - Exercises 1–6 show regions of feasible solutions....Ch. 7.2 - Prob. 6ECh. 7.2 - Use graphical methods to solve Exercises 7–12....Ch. 7.2 - Use graphical methods to solve Exercises 7–12....Ch. 7.2 - Prob. 9ECh. 7.2 - Use graphical methods to solve Exercises 7–12....Ch. 7.2 - Use graphical methods to solve Exercises 7–12....Ch. 7.2 - Use graphical methods to solve Exercises 7–12....Ch. 7.2 - Find the minimum and maximum values of (if...Ch. 7.2 - Find the minimum and maximum values of (if...Ch. 7.2 - Find the minimum and maximum values of (if...Ch. 7.2 - Find the minimum and maximum values of (if...Ch. 7.2 - Prob. 17ECh. 7.2 - 18. Find values and that maximize subject to...Ch. 7.2 - Prob. 19ECh. 7.2 - Prob. 20ECh. 7.3 - Prob. 1CPCh. 7.3 - Prob. 2CPCh. 7.3 - Prob. 3CPCh. 7.3 - Prob. 1ECh. 7.3 - Prob. 2ECh. 7.3 - Prob. 3ECh. 7.3 - Write the constraints in Exercises 1–4 as linear...Ch. 7.3 - Prob. 5ECh. 7.3 - Solve these linear programming problems, which are...Ch. 7.3 - Prob. 7ECh. 7.3 - Solve these linear programming problems, which are...Ch. 7.3 - Prob. 9ECh. 7.3 - Prob. 10ECh. 7.3 - Prob. 11ECh. 7.3 - Solve these linear programming problems, which are...Ch. 7.3 - Prob. 13ECh. 7.3 - Solve the following linear programming problems....Ch. 7.3 - Prob. 15ECh. 7.3 - Prob. 16ECh. 7.3 - Prob. 17ECh. 7.3 - Prob. 18ECh. 7.3 - Prob. 19ECh. 7.3 - Prob. 20ECh. 7.3 - Prob. 21ECh. 7.3 - Solve the following linear programming problems....Ch. 7.3 - Prob. 23ECh. 7.3 - Prob. 24ECh. 7.3 - Prob. 25ECh. 7.3 - Prob. 26ECh. 7.3 - Prob. 27ECh. 7.3 - Prob. 28ECh. 7.3 - Prob. 29ECh. 7.4 - Prob. 1CPCh. 7.4 - Prob. 2CPCh. 7.4 - Prob. 3CPCh. 7.4 - Prob. 4CPCh. 7.4 - Prob. 5CPCh. 7.4 - Prob. 6CPCh. 7.4 - Prob. 1ECh. 7.4 - Prob. 2ECh. 7.4 - In Exercises 1–4, (a) determine the number of...Ch. 7.4 - Prob. 4ECh. 7.4 - Prob. 5ECh. 7.4 - Prob. 6ECh. 7.4 - Prob. 7ECh. 7.4 - Prob. 8ECh. 7.4 - Prob. 9ECh. 7.4 - Prob. 10ECh. 7.4 - Prob. 11ECh. 7.4 - Prob. 12ECh. 7.4 - Prob. 13ECh. 7.4 - Prob. 14ECh. 7.4 - Prob. 15ECh. 7.4 - Prob. 16ECh. 7.4 - Prob. 17ECh. 7.4 - Prob. 18ECh. 7.4 - Prob. 19ECh. 7.4 - Prob. 20ECh. 7.4 - Prob. 21ECh. 7.4 - Use the simplex method to solve Exercises...Ch. 7.4 - Prob. 23ECh. 7.4 - Prob. 24ECh. 7.4 - Prob. 25ECh. 7.4 - Prob. 26ECh. 7.4 - Prob. 27ECh. 7.4 - Prob. 28ECh. 7.4 - Prob. 29ECh. 7.4 - Prob. 30ECh. 7.4 - Prob. 31ECh. 7.4 - Prob. 32ECh. 7.4 - Use the simplex method to solve Exercises...Ch. 7.4 - Prob. 34ECh. 7.4 - Use the simplex method to solve Exercises...Ch. 7.4 - Use the simplex method to solve Exercises...Ch. 7.4 - Prob. 37ECh. 7.4 - Prob. 38ECh. 7.5 - Prob. 1CPCh. 7.5 - Prob. 1ECh. 7.5 - Prob. 2ECh. 7.5 - Prob. 3ECh. 7.5 - Set up the initial simplex tableau for each of the...Ch. 7.5 - In each of the given exercises, (a) use the...Ch. 7.5 - In each of the given exercises, (a) use the...Ch. 7.5 - In each of the given exercises, (a) use the...Ch. 7.5 - Prob. 8ECh. 7.5 - Prob. 9ECh. 7.5 - Prob. 10ECh. 7.5 - Prob. 11ECh. 7.5 - Prob. 12ECh. 7.5 - Prob. 13ECh. 7.5 - Prob. 14ECh. 7.5 - Prob. 15ECh. 7.5 - Prob. 16ECh. 7.5 - Prob. 17ECh. 7.5 - Prob. 18ECh. 7.5 - Prob. 19ECh. 7.5 - Prob. 20ECh. 7.5 - Prob. 21ECh. 7.5 - Prob. 22ECh. 7.6 - Checkpoint 1 Give the transpose of each...Ch. 7.6 - Prob. 2CPCh. 7.6 - Prob. 3CPCh. 7.6 - Prob. 4CPCh. 7.6 - Prob. 5CPCh. 7.6 - Prob. 6CPCh. 7.6 - Prob. 1ECh. 7.6 - Prob. 2ECh. 7.6 - Prob. 3ECh. 7.6 - Prob. 4ECh. 7.6 - Prob. 5ECh. 7.6 - Prob. 6ECh. 7.6 - Prob. 7ECh. 7.6 - Prob. 8ECh. 7.6 - Prob. 9ECh. 7.6 - Prob. 10ECh. 7.6 - Prob. 11ECh. 7.6 - Prob. 12ECh. 7.6 - Prob. 13ECh. 7.6 - Prob. 14ECh. 7.6 - Prob. 15ECh. 7.6 - Prob. 16ECh. 7.6 - Prob. 17ECh. 7.6 - Prob. 18ECh. 7.6 - Prob. 19ECh. 7.6 - Prob. 20ECh. 7.6 - Prob. 21ECh. 7.6 - Prob. 22ECh. 7.6 - Prob. 23ECh. 7.6 - Prob. 24ECh. 7.6 - Prob. 25ECh. 7.6 - Prob. 26ECh. 7.6 - Prob. 27ECh. 7.6 - 28. Business An animal food must provide at least...Ch. 7.6 - Prob. 29ECh. 7.6 - 30. Business Joan McKee has a part-time job...Ch. 7.6 - Prob. 31ECh. 7.6 - Prob. 32ECh. 7.6 - Prob. 33ECh. 7.6 - Prob. 34ECh. 7.7 - Prob. 1CPCh. 7.7 - Prob. 2CPCh. 7.7 - Prob. 3CPCh. 7.7 - Prob. 4CPCh. 7.7 - Prob. 5CPCh. 7.7 - Prob. 1ECh. 7.7 - Prob. 2ECh. 7.7 - Prob. 3ECh. 7.7 - Prob. 4ECh. 7.7 - Prob. 5ECh. 7.7 - Prob. 6ECh. 7.7 - Prob. 7ECh. 7.7 - Prob. 8ECh. 7.7 - Prob. 9ECh. 7.7 - Prob. 10ECh. 7.7 - Prob. 11ECh. 7.7 - Prob. 12ECh. 7.7 - Prob. 13ECh. 7.7 - Prob. 14ECh. 7.7 - Prob. 15ECh. 7.7 - Prob. 16ECh. 7.7 - Prob. 17ECh. 7.7 - Prob. 18ECh. 7.7 - Prob. 19ECh. 7.7 - Prob. 20ECh. 7.7 - Prob. 21ECh. 7.7 - Prob. 22ECh. 7.7 - Prob. 23ECh. 7.7 - Prob. 24ECh. 7.7 - Prob. 25ECh. 7.7 - Prob. 26ECh. 7.7 - Prob. 27ECh. 7.7 - Prob. 28ECh. 7.7 - Prob. 29ECh. 7.7 - Prob. 30ECh. 7.7 - Prob. 31ECh. 7.7 - Prob. 32ECh. 7.7 - Prob. 33ECh. 7.7 - Prob. 34ECh. 7.7 - Prob. 35ECh. 7.7 - Use the two-stage method to solve Exercises 33–40....Ch. 7.7 - Prob. 37ECh. 7.7 - Prob. 38ECh. 7.7 - Prob. 39ECh. 7.7 - Prob. 40ECh. 7.7 - Prob. 41ECh. 7.7 - Prob. 42ECh. 7.7 - Prob. 43ECh. 7 - Prob. 1CECh. 7 - 2. Consider preparing a stir-fry using beef, oil,...Ch. 7 - Prob. EPCh. 7 - Prob. 1RECh. 7 - Prob. 2RECh. 7 - Graph each of the given linear inequalities. 3. Ch. 7 - Prob. 4RECh. 7 - Prob. 5RECh. 7 - Prob. 6RECh. 7 - Prob. 7RECh. 7 - Prob. 8RECh. 7 - Prob. 9RECh. 7 - Prob. 10RECh. 7 - Prob. 11RECh. 7 - Prob. 12RECh. 7 - Prob. 13RECh. 7 - Use the graphical method to solve Exercises...Ch. 7 - Use the graphical method to solve Exercises...Ch. 7 - Prob. 16RECh. 7 - Prob. 17RECh. 7 - Prob. 18RECh. 7 - 19. Finance The BlackRock Equity Dividend Fund...Ch. 7 - Prob. 20RECh. 7 - Prob. 21RECh. 7 - Prob. 22RECh. 7 - Prob. 23RECh. 7 - Prob. 24RECh. 7 - Prob. 25RECh. 7 - Prob. 26RECh. 7 - Prob. 27RECh. 7 - Prob. 28RECh. 7 - Prob. 29RECh. 7 - Prob. 30RECh. 7 - Prob. 31RECh. 7 - Prob. 32RECh. 7 - Prob. 33RECh. 7 - Prob. 34RECh. 7 - Prob. 35RECh. 7 - Prob. 36RECh. 7 - 37. When is it necessary to use the simplex method...Ch. 7 - Prob. 38RECh. 7 - 39. What kind of problem can be solved with the...Ch. 7 - 40. In solving a linear programming problem, you...Ch. 7 - Prob. 41RECh. 7 - Prob. 42RECh. 7 - Prob. 43RECh. 7 - Prob. 44RECh. 7 - Prob. 45RECh. 7 - Use the method of duals to solve these...Ch. 7 - Prob. 47RECh. 7 - Prob. 48RECh. 7 - Prob. 49RECh. 7 - Prob. 50RECh. 7 - Prob. 51RECh. 7 - Prob. 52RECh. 7 - Use the two-stage method to solve these...Ch. 7 - Prob. 54RECh. 7 - Prob. 55RECh. 7 - Prob. 56RECh. 7 - Business Solve the following maximization...Ch. 7 - Prob. 58RECh. 7 - Business Solve the following maximization...Ch. 7 - Prob. 60RECh. 7 - Prob. 61RECh. 7 - Business Solve the following minimization...Ch. 7 - Business Solve these mixed-constraint...Ch. 7 - Business Solve these mixed-constraint...
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