Solve the following l.p.p using simplex method

Maximum z=10x+20y
Subject to 3x+5y<90
6×+3y<72
X,y>0

graph first constraint line

(0, 18) and (30,0)
graph second constraint line (really 2 x+ y < 24)
(0, 24) and (12,0)
look at all interesting corners
(0,18) and (30/7, 15.43) and (12,0)
compute 10 x + 20 y at each of those corners
pick the big one (it is the middle one I bet :)

To solve the linear programming problem (LPP) using the simplex method, we will follow these steps:

Step 1: Convert the problem into standard form by introducing slack variables:
3x + 5y + s1 = 90
6x + 3y + s2 = 72

Step 2: Convert the objective function into minimization form:
Minimize -z = -10x - 20y

Step 3: Create the initial simplex tableau:
Max z | x | y | s1 | s2 | RHS
-----------------------------------------
0 | -10 | -20 | 0 | 0 | 0
-----------------------------------------
s1 | 3 | 5 | 1 | 0 | 90
-----------------------------------------
s2 | 6 | 3 | 0 | 1 | 72

Step 4: Select the pivot column:
The most negative coefficient in the objective row is -20, corresponding to variable y. So, we select variable y as the pivot column.

Step 5: Select the pivot row:
To find the pivot row, calculate the ratio of the RHS value to the corresponding coefficient for each constraint. The smallest positive ratio will determine the pivot row.
For constraint 1: 90/5 = 18
For constraint 2: 72/3 = 24

The smallest ratio is 18, corresponding to constraint 1. So, we select constraint 1 as the pivot row.

Step 6: Perform row operations to make the pivot element equal to 1 and make all other elements in the pivot column zero:
Divide the pivot row by the pivot element (5) to make the pivot element equal to 1:
s1 = (1/5)s1

Perform the row operations to make other elements in the pivot column zero:
R2 = R2 - 3R1
s2 = s2 - 6s1

After the row operations, the simplex tableau becomes:
Max z | x | y | s1 | s2 | RHS
-----------------------------------------
0 | -10 | 0 | -4 | 0 | -90
-----------------------------------------
s1 | 3 | 0 | 1/5 | 0 | 18
-----------------------------------------
s2 | 6 | 0 | -6/5| 1 | 6

Step 7: Perform iterations until all the coefficients in the objective row are non-negative.

Iteration 1:
Pivot column: x (as it has the most negative coefficient in the objective row)
Pivot row: s2 (as it has the smallest positive ratio)

Perform row operations to make the pivot element equal to 1 and other elements in the pivot column zero:
s2 = (1/6)s2

R1 = R1 - 3R2
s1 = s1 + (6/5)s2

After the row operations, the simplex tableau becomes:
Max z | x | y | s1 | s2 | RHS
------------------------------------------
0 | 0 | 0 | 16/5| -1 | -84
------------------------------------------
s1 | 0 | 0 | 7/5 | -1/6| 22
------------------------------------------
s2 | 1 | 0 | -1 | 1/6 | 1

As all coefficients in the objective row are non-negative, we have reached the optimal solution.
The optimal solution is:
x = 1
y = 0
z = 0

The maximum value of z is 0 when x = 1 and y = 0.

To solve the given linear programming problem (LPP) using the simplex method, we need to follow these steps:

Step 1: Set up the initial tableau
We first need to convert the given LPP into standard form by introducing slack variables. The standard form for a maximization problem is as follows:

Maximize z = 10x + 20y
Subject to:
3x + 5y + s1 = 90
6x + 3y + s2 = 72
x, y, s1, s2 ≥ 0

Now, we construct the initial tableau by incorporating the objective function coefficients, the constraint coefficients, and the slack variables:

| -3x - 5y - s1 + Z 0 |
|--------------------------|
| 3x + 5y + s1 = 90 |
| 6x + 3y = 72 |
| x, y ≥ 0 |

Step 2: Select the pivot element
Next, we determine the pivot column, which is the most negative entry in the bottom row (excluding the Z column). If there are multiple negative values, choose the most negative one. In this example, the most negative value is -5.

Step 3: Determine the pivot row
To find the pivot row, we divide the right-hand side (RHS) values by their corresponding coefficients in the pivot column. The smallest non-negative ratio indicates the pivot row. In this case, the ratios are 90/5 = 18 and 72/3 = 24. Therefore, the pivot row is the one with the smallest ratio, which is the first row.

Step 4: Update the tableau
Now, we perform row operations to make the pivot element (in this case -5) equal to 1. We divide the entire pivot row by -5:

| 0x + y + (1/5)s1 - Z - 0 |
|---------------------------------|
| 3x + 5y + s1 = 90 |
| 6x + 3y = 72 |
| x, y ≥ 0 |

Using row operations, we make the elements in the pivot column equal to zero:

| 0x + y + (1/5)s1 - Z + (3/5)s2 - 18 |
|-------------------------------------------------|
| 1x + (5/3)y + (1/3)s1 - 30 |
|-------------------------------------------------|
| 0x - (7/3)y + (2/3)s1 + 18 |
|-------------------------------------------------|
| x, y ≥ 0 |

Step 5: Repeat steps 2-4
We repeat steps 2-4 until there are no more negative values in the bottom row. In each iteration, we select a new pivot column, determine the pivot row, and update the tableau accordingly.

Step 6: Find the optimal solution
If there are no more negative values in the bottom row, we have reached the optimal solution. In this case, the optimal solution is found when the objective function (Z) value in the bottom row is zero. We can read the optimal values of x, y, and the objective function from the tableau.

Thus, the optimal solution is:
z = 60 (maximum value of the objective function)
x = 10
y = 0

Therefore, the maximum value of z is 60, and it occurs when x = 10 and y = 0.