Maximize $$$7 x + 2 y$$$, subject to $$$\begin{cases} 3 x + 5 y \geq 20 \\ 3 x + y \leq 16 \\ - 2 x + y \leq 1 \\ x \geq 0 \\ y \geq 0 \end{cases}$$$

Maximize $$$Z = 7 x + 2 y$$$, subject to $$$\begin{cases} 3 x + 5 y \geq 20 \\ 3 x + y \leq 16 \\ - 2 x + y \leq 1 \\ x \geq 0 \\ y \geq 0 \end{cases}.$$$

The problem in the canonical form can be written as follows:

Add variables (slack or surplus) to turn all the inequalities into equalities:

Since we don't have a basis, add an artificial variable:

Penalize the artificial variable in the objective function:

Write down the simplex tableau:

Make the Z-row consistent with the rest of the tableau.

Subtract row $$$2$$$ multiplied by $$$M$$$ from row $$$1$$$: $$$R_{1} = R_{1} - M R_{2}$$$.

The entering variable is $$$y$$$, because it has the most negative coefficient $$$- 5 M - 2$$$ in the Z-row.

The leaving variable is $$$S_{3}$$$, because it has the smallest ratio.

Add row $$$4$$$ multiplied by $$$5 M + 2$$$ to row $$$1$$$: $$$R_{1} = R_{1} + \left(5 M + 2\right) R_{4}$$$.

Subtract row $$$4$$$ multiplied by $$$5$$$ from row $$$2$$$: $$$R_{2} = R_{2} - 5 R_{4}$$$.

Subtract row $$$4$$$ from row $$$3$$$: $$$R_{3} = R_{3} - R_{4}$$$.

The entering variable is $$$x$$$, because it has the most negative coefficient $$$- 13 M - 11$$$ in the Z-row.

The leaving variable is $$$Y_{1}$$$, because it has the smallest ratio.

Divide row $$$1$$$ by $$$13$$$: $$$R_{1} = \frac{R_{1}}{13}$$$.

Add row $$$2$$$ multiplied by $$$13 M + 11$$$ to row $$$1$$$: $$$R_{1} = R_{1} + \left(13 M + 11\right) R_{2}$$$.

Subtract row $$$2$$$ multiplied by $$$5$$$ from row $$$3$$$: $$$R_{3} = R_{3} - 5 R_{2}$$$.

Add row $$$2$$$ multiplied by $$$2$$$ to row $$$4$$$: $$$R_{4} = R_{4} + 2 R_{2}$$$.

The entering variable is $$$S_{3}$$$, because it has the most negative coefficient $$$- \frac{29}{13}$$$ in the Z-row.

The leaving variable is $$$S_{2}$$$, because it has the smallest ratio.

Multiply row $$$2$$$ by $$$\frac{13}{12}$$$: $$$R_{2} = \frac{13 R_{2}}{12}$$$.

Add row $$$3$$$ multiplied by $$$\frac{29}{13}$$$ to row $$$1$$$: $$$R_{1} = R_{1} + \frac{29 R_{3}}{13}$$$.

Add row $$$3$$$ multiplied by $$$\frac{5}{13}$$$ to row $$$2$$$: $$$R_{2} = R_{2} + \frac{5 R_{3}}{13}$$$.

Subtract row $$$3$$$ multiplied by $$$\frac{3}{13}$$$ from row $$$4$$$: $$$R_{4} = R_{4} - \frac{3 R_{3}}{13}$$$.

None of the Z-row coefficients are negative.

The optimum is reached.

The following solution is obtained: $$$\left(x, y, S_{1}, S_{2}, S_{3}, Y_{1}\right) = \left(5, 1, 0, 0, 10, 0\right)$$$.

$$$Z = 37$$$A is achieved at $$$\left(x, y\right) = \left(5, 1\right)$$$A.