By Simplex method maximize [6] F = 15x 1 + 10x 2 subject to 2x 1 + x 2 10 , x 1 + 3x 2 10 ; x 1, x 2 0 .

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Solution: Maximize $F=15x_1+10x_2$ subject to constraints

Step 1: Convert the problem into standard form

The given constraints are:

1. $2x_1+x_2\le 10$
2. $x_1+3x_2\le 10$
3. $x_1,x_2\ge 0$

To apply the Simplex method, we convert the inequalities into equations by introducing slack variables $s_1$ and $s_2$:

1. $2x_1+x_2+s_1=10$
2. $x_1+3x_2+s_2=10$

The objective function remains: $$F=15x_1+10x_2+0s_1+0s_2$$

Step 2: Construct the initial Simplex tableau

The initial tableau is written as follows:

$$\begin{array}{cccccc}& x_1& x_2& s_1& s_2& RHS\\ s_1& 2& 1& 1& 0& 10\\ s_2& 1& 3& 0& 1& 10\\ F& -15& -10& 0& 0& 0\end{array}$$

Step 3: Identify the pivot column

The pivot column is selected based on the most negative coefficient in the $F$ row (objective row). Here, $-15$ (for $x_1$) is the most negative, so $x_1$ is the pivot column.

Step 4: Determine the pivot row

The pivot row is found by calculating the minimum ratio of the RHS to the pivot column coefficients (ignoring non-positive values):

• For $s_1$: $10/2=5$
• For $s_2$: $10/1=10$

The smallest ratio is 5, corresponding to the $s_1$ row. Thus, $s_1$ is the pivot row.

Step 5: Perform row operations to update the tableau

We perform the following operations to make the pivot element $1$ and eliminate other entries in the pivot column:

1. Divide the pivot row ($s_1$) by $2$: $$\text{New }{s}_1\text{ row}=\frac{1}{2}\times [2,1,1,0,10]=[1,0.5,0.5,0,5]$$

2. Eliminate $x_1$ from the $s_2$ row: $$\text{New }{s}_2\text{ row}=[1,3,0,1,10]-1\times [1,0.5,0.5,0,5]=[0,2.5,-0.5,1,5]$$

3. Eliminate $x_1$ from the $F$ row: $$\text{New }F\text{ row}=[-15,-10,0,0,0]+15\times [1,0.5,0.5,0,5]=[0,-2.5,7.5,0,75]$$

The updated tableau is:

$$\begin{array}{cccccc}& x_1& x_2& s_1& s_2& RHS\\ x_1& 1& 0.5& 0.5& 0& 5\\ s_2& 0& 2.5& -0.5& 1& 5\\ F& 0& -2.5& 7.5& 0& 75\end{array}$$

Step 6: Check for optimality

The $F$ row now has no negative coefficients, indicating that the optimal solution has been reached.

Step 7: Extract the solution

From the final tableau:

• $x_1=5$ (basic variable in the first row)
• $x_2=0$ (since its column has no basic variable)
• $s_1=0$ (since $x_1$ is in the basis)
• $s_2=5$ (basic variable in the second row)

The maximum value of $F$ is 75.

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Verification of the solution

Substitute $x_1=5$ and $x_2=0$ into the original constraints:

1. $2(5)+0=10\le 10$ ✓
2. $5+3(0)=5\le 10$ ✓

Both constraints are satisfied, and the solution is feasible.

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Final Answer

The maximum value of the objective function is achieved at: $${\mathbf{x}}_1=5,{\mathbf{x}}_2=0$$ with the maximum value of $F=75$.