consider the following simplex tableau being used to maximize z. the current solution is not optimal and…

Answer:

(A) (x_{2}) (B)

Row	Variable	Ratio as Decimal
Row 1	(s_{1})	(65.33)
Row 2	(s_{2})	(22.00)
Row 3	(s_{3})	(DNE)
(C) (s_{2})

Explanation:

Step1: Determine the entering variable

In the simplex method for maximization, we look at the bottom - row (the (Z) - row excluding the (Z) - value). The most positive coefficient in the non - basic variable columns ((x_1,x_2,x_3)) determines the entering variable. The coefficients of (x_1,x_2,x_3) in the (Z) - row are (- 3,1,0). Since (1) (the coefficient of (x_2)) is the most positive, the entering variable is (x_2).

Step2: Calculate the ratios

The ratio for a row (i) is calculated as (\frac{\text{Right - hand side value of row }i}{\text{Entry in the entering variable column of row }i})

• For Row 1: The right - hand side value is (196) and the entry in the (x_2) column is (3). The ratio is (\frac{196}{3}\approx65.33), and the variable in Row 1 is (s_{1}) (the slack variable corresponding to the first constraint)
• For Row 2: The right - hand side value is (66) and the entry in the (x_2) column is (3). The ratio is (\frac{66}{3}=22.00), and the variable in Row 2 is (s_{2}) (the slack variable corresponding to the second constraint)
• For Row 3: The entry in the (x_2) column is (2), but we consider non - negative ratios. Since we are maximizing, we ignore non - positive ratios. The right - hand side value is (151) and the entry in the (x_2) column is (2) (but we use the formula (\frac{\text{RHS}}{\text{Entry in entering variable column}}). However, if we consider the standard simplex ratio calculation for maximization (non - negative ratios), for the third row, if we assume the formula (\frac{151}{2} = 75.5), but looking at the simplex tableau structure (the original problem's constraints), we should use the correct columns. Wait, no, re - checking: The correct formula for the ratio is (\text{Ratio}=\frac{b_i}{a_{ij}}) where (b_i) is the right - hand side of the (i) - th row and (a_{ij}) is the entry in the (j) - th column (entering variable column) of the (i) - th row. For Row 3: The entry in the (x_2) column is (2) (assuming it's a typo in the problem statement's tableau, if we follow the ratio rule for maximization (non - negative ratios for non - negative (a_{ij})). But actually, in the simplex tableau for maximization, we calculate (\frac{b_i}{a_{ij}}) for (a_{ij}>0). The right - hand side of Row 3 is (151) and the entry in the (x_2) column (entering variable) is (2). But wait, no, looking at the tableau: The first three rows are constraint rows. The right - hand side values are (196) (Row 1), (66) (Row 2), (151) (Row 3). The entries in the (x_2) column (entering variable) for Row 1: (3), Row 2: (3), Row 3: (2). But in maximization simplex, we calculate (\frac{b_i}{a_{ij}}) for (a_{ij}>0). However, if we consider the standard ratio test (to find the smallest non - negative ratio), but actually, we made a mistake above. Wait, no, re - looking at the simplex tableau: The bottom row ( (Z) - row) has coefficients for (x_1,x_2,x_3,s_1,s_2,s_3) as (-3,1,0,0, - 2,0). The entering variable is (x_2) (since (1) is the most positive in the non - basic variable columns ((x_1,x_2,x_3))). For the ratio calculation: Row 1: (\frac{196}{3}\approx65.33) (variable (s_1)) Row 2: (\frac{66}{3} = 22.00) (variable (s_2)) Row 3: The entry in the (x_2) column is (2), but if we assume the correct formula (\frac{\text{RHS}}{a_{ij}}), but in the context of the simplex method (maximization), we consider non - negative ratios. However, if we look at the original problem (assuming the tableau is set up correctly), the third row's right - hand side is (151) and (a_{ij} = 2) (for (x_2)), but actually, no, we made a mistake. Wait, no, the first three rows are: Row 1: (6x_1+3x_2 + 0x_3+s_1+0s_2+0s_3=196) Row 2: (7x_1 + 3x_2+0x_3+0s_1 + 2s_2+1s_3=66) Row 3: (3x_1+2x_2+1x_3+0s_1+3s_2+0s_3=151) The ratio for Row 1: (\frac{196}{3}\approx65.33) ( (s_1) is the basic variable) The ratio for Row 2: (\frac{66}{3}=22.00) ( (s_2) is the basic variable) The ratio for Row 3: (\frac{151}{2}=75.5), but wait, no, in the simplex method, when we have (a_{ij}) (entry in entering variable column) positive, we calculate the ratio. But if we consider the standard ratio test (to find the smallest non - negative ratio). The smallest non - negative ratio is (22.00) (from Row 2)

Step3: Determine the exiting variable

The exiting variable is the basic variable corresponding to the row with the smallest non - negative ratio. Since the smallest non - negative ratio ((22.00)) is in Row 2 and the basic variable in Row 2 is (s_{2}), the exiting variable is (s_{2})