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.

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:

$\underset{\_}{\frac{1}{6}}$ Cup of feta cheese, $\underset{\_}{4\frac{1}{4}}$ cup of lettuce, $\underset{\_}{\frac{1}{8}}$ cup of salad dressing, $\underset{\_}{\frac{7}{8}}$ 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	$6\text{oz}=170\text{g}$
Egg	$1\text{egg}=61\text{g}$
Feta Cheese	$\frac{1}{4}\text{cup}=38\text{g}$
Lettuce	$\frac{1}{2}\text{cup}=28\text{g}$
Milk	$1\text{cup}=244\text{g}$
Oil	$1\text{Tbsp}=13.5\text{g}$
Onion	$1\text{Onion}=110\text{g}$
Salad Dressing	$1\text{cup}=250\text{g}$
Soy Sauce	$1\text{Tbsp}=18\text{g}$
spinach	$1\text{cup}=180\text{g}$
Tomato	$1\text{tomato}=123\text{g}$

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 $x_1$ be the number of 100-gram units of feta cheese, take $x_2$ be the number of 100-gram units of lettuce, and $x_3$ be the number of 100-gram units of salad dressing and take $x_4$ 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:

$z=4.09x_1+2.37x_2+2.5x_3+4.64x_4$

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

$263x_1+14x_2+448.8x_3+21x_4<260$

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

$492.5x_1+36x_2+0x_3+5x_4>210$

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

$10.33x_1+1.62x_2+0x_3+.85x_4>6$

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

$x_1+x_2+x_3+x_4<4$

Consider the conversion table,

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

Food	Serving size
Beef	$6\text{oz}=170\text{g}$
Egg	$1\text{egg}=61\text{g}$
Feta Cheese	$\frac{1}{4}\text{cup}=38\text{g}$
Lettuce	$\frac{1}{2}\text{cup}=28\text{g}$
Milk	$1\text{cup}=244\text{g}$
Oil	$1\text{Tbsp}=13.5\text{g}$
Onion	$1\text{Onion}=110\text{g}$
Salad Dressing	$1\text{cup}=250\text{g}$
Soy Sauce	$1\text{Tbsp}=18\text{g}$
spinach	$1\text{cup}=180\text{g}$
Tomato	$1\text{tomato}=123\text{g}$

Salad dressing is at least 2 tablespoons which is $\frac{1}{8}$ cup from the table it converts into gram which is:

$\begin{array}{c}1\text{cup}=250\text{g}\\ \frac{1}{8}\text{cup}\text{=}\frac{250}{8}\\ \approx 31.25\text{g}\end{array}$

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

$\begin{array}{c}100\text{gram}=1\text{unit}\\ \text{1}\text{gram}=\frac{1}{100}\end{array}$

And,

$\begin{array}{c}31.25\text{gram}=\frac{1}{100}\times 31.25\\ =.3125\end{array}$

So, the constraints for salad dressing is:

$x_3\le .3125$

Therefore, formulate the problem which is:

Maximize,

$z=4.09x_1+2.37x_2+2.5x_3+4.64x_4$

Subject to,

$\begin{array}{c}263x_1+14x_2+448.8x_3+21x_4<260\\ 492.5x_1+36x_2+0x_3+5x_4>210\\ 10.33x_1+1.62x_2+0x_3+.85x_4>6\\ x_1+x_2+x_3+x_4<4\end{array}$

And,

$\begin{array}{c}{x}_3\le .3125\\ x_1\ge 0,x_2\ge 0,x_3\ge 0,and{x}_4\ge 0\end{array}$

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:

$\begin{array}{c}263x_1+14x_2+448.8x_3+21x_4+s_1=260\\ 492.5x_1+36x_2+0x_3+5x_4-s_2=210\\ 10.33x_1+1.62x_2+0x_3+.85x_4-s_3=6\\ x_1+x_2+x_3+x_4+s_4=4\end{array}$

And,

$\begin{array}{c}{x}_3+s_5=.3125,\\ x_1\ge 0,x_2\ge 0,x_3\ge 0,\text{ and }{x}_4\ge 0\end{array}$

Where $s_1,s_2,s_3,s_4$ are greater to zero and $s_5\ge 0$

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

Maximize,

$z=4.09x_1+2.37x_2+2.5x_3+4.64x_4$

Subject to,

$\begin{array}{c}263x_1+14x_2+448.8x_3+21x_4+s_1=260\\ 492.5x_1+36x_2+0x_3+5x_4-s_2=210\\ 10.33x_1+1.62x_2+0x_3+.85x_4-s_3=6\\ x_1+x_2+x_3+x_4+s_4=4\end{array}$

And,

$\begin{array}{c}{x}_3+s_5=0.3125\\ x_1\ge 0,x_2\ge 0,x_3\ge 0,x_4\ge 0\\ s_1>0,s_2>0,s_3>0,s_4>0,s_5\ge 0\end{array}$

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:

$-4.09x_1-2.37x_2-2.5x_3-4.64x_4+z=0$

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

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}263& 14& 448.8& 21& 1& 0& 0& 0& 0& 260\\ 492.5& 36& 0& 5& 0& -1& 0& 0& 0& 210\\ 10.33& 1.62& 0& .85& 0& 0& -1& 0& 0& 6\\ 1& 1& 1& 1& 0& 0& 0& 1& 0& 4\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ -4.09& -2.37& -2.5& -4.64& 0& 0& 0& 0& 0& 0\end{array}\right]\end{array}$

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 $x_1,x_2,x_3,x_4$ and the basic variables are $s_1,s_2,s_3,s_4,s_5$.

Here,

$x_1=0,x_2=0,x_3=0,x_4=0$, and

$s_1=260,s_2=-210,s_3=-6,s_4=4,s_5=-.3125$

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:

$x_1,x_2,x_3,x_4,s_1,s_2,s_3,s_4,s_5,z$.

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 $210$.

So, take $x_1$ column as pivot column.

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

First, $260÷263=.98$.

Second, $210÷492.5=.426$.

Third, $6÷10.33=.58$.

Fourth, $4÷1=4$.

Fifth, $.3125÷0=\infty$.

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.

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}263& 14& 448.8& 21& 1& 0& 0& 0& 0& 260\\ \boxed{492.5}& 36& 0& 5& 0& -1& 0& 0& 0& 210\\ 10.33& 1.62& 0& .85& 0& 0& -1& 0& 0& 6\\ 1& 1& 1& 1& 0& 0& 0& 1& 0& 4\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ -4.09& -2.37& -2.5& -4.64& 0& 0& 0& 0& 0& 0\end{array}\right]\end{array}$

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 $\frac{1}{492.5}$ times itself which is:

$R_2\to \frac{1}{492.5}{R}_2$

Then,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}263& 14& 448.8& 21& 1& 0& 0& 0& 0& 260\\ 1& .073& 0& .010& 0& -.002& 0& 0& 0& .426\\ 10.33& 1.62& 0& .85& 0& 0& -1& 0& 0& 6\\ 1& 1& 1& 1& 0& 0& 0& 1& 0& 4\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ -4.09& -2.37& -2.5& -4.64& 0& 0& 0& 0& 0& 0\end{array}\right]\end{array}$

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

$R_1\to R_1-263R_2$

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

$R_3\to R_3-10.33R_2$

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

$R_4\to R_4-R_2$

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

$R_6\to R_6+4.09R_2$

Then,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& -5.19& 448.8& 18.37& 1& .526& 0& 0& 0& 147.96\\ 1& .073& 0& .010& 0& -.002& 0& 0& 0& .426\\ 0& .865& 0& .746& 0& .020& -1& 0& 0& 1.59\\ 0& .927& 1& .99& 0& .002& 0& 1& 0& 3.57\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& -2.07& -2.5& -4.59& 0& -.008& 0& 0& 0& 1.742\end{array}\right]\end{array}$

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

$x_1=.426,x_2=0,x_3=0,and{x}_4=0$

$s_1=147.96,s_2=0,s_3=-1.59,s_4=3.57,and{s}_5=-.3125$

In the infeasible solution as $s_5$ 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 $x_3$ column as pivot column.

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

First, $147.96÷448.8=.33$,

Second, $.426÷0=\infty$,

Third, $1.59÷0=\infty$,

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.

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& -5.19& 448.8& 18.37& 1& .526& 0& 0& 0& 147.96\\ 1& .073& 0& .010& 0& -.002& 0& 0& 0& .426\\ 0& .865& 0& .746& 0& .020& -1& 0& 0& 1.59\\ 0& .927& 1& .99& 0& .002& 0& 1& 0& 3.57\\ 0& 0& \boxed{1}& 0& 0& 0& 0& 0& -1& .3125\\ 0& -2.07& -2.5& -4.59& 0& -.008& 0& 0& 0& 1.742\end{array}\right]\end{array}$

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:

$R_1\to R_1-448.8R_5$

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

$R_4\to R_4-R_5$

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

$R_6\to R_6+2.5R_5$

Then,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& -5.19& 0& 18.37& 1& .526& 0& 0& 448.8& 7.60\\ 1& .073& 0& .010& 0& -.002& 0& 0& 0& .426\\ 0& .865& 0& .746& 0& .020& -1& 0& 0& 1.59\\ 0& .927& 0& .99& 0& .002& 0& 1& -1& 3.26\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& -2.07& 0& -4.59& 0& -.008& 0& 0& -2.5& 2.52\end{array}\right]\end{array}$

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

$x_1=.426,x_2=0,x_3=.3125,x_4=0$,

And, $s_1=7.60,s_2=0,s_3=-1.59,s_4=3.26,s_5=0$.

In the infeasible solution, $s_3$ 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 $1.59$.

So, take $x_2$ 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, $.3125÷0=\infty$.

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.

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& -5.19& 0& 18.37& 1& .526& 0& 0& 448.8& 7.60\\ 1& .073& 0& .010& 0& -.002& 0& 0& 0& .426\\ 0& \boxed{.865}& 0& .746& 0& .020& -1& 0& 0& 1.59\\ 0& .927& 0& .99& 0& .002& 0& 1& -1& 3.26\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& -2.07& 0& -4.59& 0& -.008& 0& 0& -2.5& 2.52\end{array}\right]\end{array}$

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 $\frac{1}{.865}$ times itself which is:

$R_3\to \frac{1}{.865}{R}_3$

Then,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& -5.19& 0& 18.37& 1& .526& 0& 0& 448.8& 7.60\\ 1& .073& 0& .010& 0& -.002& 0& 0& 0& .426\\ 0& 1& 0& .861& 0& .024& -1.156& 0& 0& 1.844\\ 0& .927& 0& .99& 0& .002& 0& 1& -1& 3.26\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& -2.07& 0& -4.59& 0& -.008& 0& 0& -2.5& 2.52\end{array}\right]\end{array}$

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

$R_1\to R_1+5.19R_3$

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

$R_2\to R_2-.073R_3$

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

$R_4\to R_4-.927R_3$

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

$R_6\to R_6+2.07R_3$

Then,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& 0& 0& 22.83& 1& .660& -6.04& 0& 448.8& 17.244\\ 1& 0& 0& -.0528& 0& -.0038& .0845& 0& 0& .2916\\ 0& 1& 0& .861& 0& .024& -1.156& 0& 0& 1.844\\ 0& 0& 0& .191& 0& -.020& 1.071& 1& 1& 1.551\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& 0& 0& -2.80& 0& .0416& -2.39& 0& -2.5& 6.337\end{array}\right]\end{array}$

Here,

$x_1=.2916,x_2=1.844,x_3=.3125,x_4=0$,

And, $,s_1=17.244,s_2=0,s_3=0,s_4=1.551,s_5=0$.

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 $x_4$.

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, $.3125÷0=\infty$.

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

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& 0& 0& \boxed{22.83}& 1& .660& -6.04& 0& 448.8& 17.244\\ 1& 0& 0& -.0528& 0& -.0038& .0845& 0& 0& .2916\\ 0& 1& 0& .861& 0& .024& -1.156& 0& 0& 1.844\\ 0& 0& 0& .191& 0& -.020& 1.071& 1& 1& 1.551\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& 0& 0& -2.80& 0& .0416& -2.39& 0& -2.5& 6.337\end{array}\right]\end{array}$

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 $\frac{1}{22.83}$ times of itself which is:

$R_1\to \frac{1}{22.83}{R}_1$

Then,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& 0& 0& 1& .043& .0289& -.2646& 0& 19.65& .7553\\ 1& 0& 0& -.0528& 0& -.0038& .0845& 0& 0& .2916\\ 0& 1& 0& .861& 0& .024& -1.156& 0& 0& 1.844\\ 0& 0& 0& .191& 0& -.020& 1.071& 1& 1& 1.551\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& 0& 0& -2.80& 0& .0416& -2.39& 0& -2.5& 6.337\end{array}\right]\end{array}$

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

$R_2\to R_2+.0528R_1$

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

$R_3\to R_3-.861R_1$

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

$R_4\to R_4-.191R_1$

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

$R_6\to R_6+2.80R_1$

Then,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& 0& 0& 1& .043& .0289& -.2646& 0& 19.65& .7553\\ 1& 0& 0& 0& .0023& -.0023& .0705& 0& 1.038& .3315\\ 0& 1& 0& 0& -.0377& -.0007& -.9283& 0& -16.935& 1.193\\ 0& 0& 0& 0& -.0084& -.026& 1.122& 1& -2.760& 1.406\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& 0& 0& 0& .1204& .1225& -3.12& 0& 52.52& 8.451\end{array}\right]\end{array}$

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 $s_3$.

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, $.3125÷0=\infty$.

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,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& 0& 0& 1& .043& .0289& -.2646& 0& 19.65& .7553\\ 1& 0& 0& 0& .0023& -.0023& .0705& 0& 1.038& .3315\\ 0& 1& 0& 0& -.0377& -.0007& -.9283& 0& -16.935& 1.193\\ 0& 0& 0& 0& -.0084& -.026& \boxed{1.122}& 1& -2.760& 1.406\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& 0& 0& 0& .1204& .1225& -3.12& 0& 52.52& 8.451\end{array}\right]\end{array}$

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 $\frac{1}{1.122}$ times of itself which is:

$R_4\to \frac{1}{1.122}{R}_4$

Then,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& 0& 0& 1& .043& .0289& -.2646& 0& 19.65& .7553\\ 1& 0& 0& 0& .0023& -.0023& .0705& 0& 1.038& .3315\\ 0& 1& 0& 0& -.0377& -.0007& -.9283& 0& -16.935& 1.193\\ 0& 0& 0& 0& -.0075& -.0232& 1& .891& -2.46& 1.253\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& 0& 0& 0& .1204& .1225& -3.12& 0& 52.52& 8.451\end{array}\right]\end{array}$

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

$R_1\to R_1+.2646R_4$

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

$R_2\to R_2-.0705R_4$

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

$R_3\to R_3+.9283R_4$

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

$R_6\to R_6+3.12R_4$

Then,

$\begin{array}{c}{x}_1{x}_2{x}_3{x}_4{s}_1{s}_2{s}_3{s}_4{s}_5\\ \left[\begin{array}{cccccccccc}0& 0& 0& 1& .0418& .0228& 0& .2357& 19.006& 1.087\\ 1& 0& 0& 0& .0028& -.0006& 0& -.0629& 1.2119& .243\\ 0& 1& 0& 0& -.0447& -.0222& 0& .8271& -19.2187& 2.375\\ 0& 0& 0& 0& -.0075& -.0232& 1& .891& -2.46& 1.253\\ 0& 0& 1& 0& 0& 0& 0& 0& -1& .3125\\ 0& 0& 0& 0& .0998& .0507& 0& 2.797& 45.099& 12.41\end{array}\right]\end{array}$

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:

$x_1=.243,x_2=2.375,x_3=.3125,x_4=1.087,s_1=17.244,s_2=0,s_3=0,s_4=1.551,s_5=0$

So, the solution of the maximum value is:

$\begin{array}{c}z=4.09x_1+2.37x_2+2.5x_3+4.64x_4\\ =4.09\left(.243\right)+2.37\left(2.37\right)+2.5\left(.3125\right)+4.64\left(1.087\right)\\ =.9938+5.569+.78125+5.04\\ =12.41\end{array}$

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	$\frac{1}{4}\text{cup}=38\text{g}$
Lettuce	$\frac{1}{2}\text{cup}=28\text{g}$
Salad Dressing	$1\text{cup}=250\text{g}$
Tomato	$1\text{tomato}=123\text{g}$

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

$\begin{array}{c}38\text{g}=\frac{1}{4}\text{cup}\\ \text{1=}\frac{1}{4\times 38}\\ 24.3=\frac{1}{4\times 38}\times 24.3\\ =\frac{1}{6}\text{cup}\end{array}$

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

$\begin{array}{c}28\text{g}=\frac{1}{2}\text{cup}\\ \text{1=}\frac{1}{2\times 28}\\ 237.5=\frac{1}{2\times 28}\times 237.5\\ =\frac{17}{4}\text{cup}\approx \text{4}\frac{1}{4}\end{array}$

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

$\begin{array}{c}250\text{g}=1\text{cup}\\ \text{1=}\frac{1}{250}\\ 31.25=\frac{1}{250}\times 31.25\\ =\frac{1}{8}\text{cup}\end{array}$

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

$\begin{array}{c}123\text{g}=1\text{tomato}\\ \text{1}\text{g }=\frac{1}{123}\\ 108.7=\frac{1}{123}\times 108.7\\ =\frac{7}{8}\text{of 1}\text{tomato}\end{array}$