Solve the following linear programming problem by Simplex method to maximize P = 50x + 60y subject to 3x + 4y 36, 9x + 4y 60, x, y 0

Solution by Simplex Method

Step 1: Convert the inequalities into equations by introducing slack variables $s_1$ and $s_2$.

The given linear programming problem is: Maximize $P=50x+60y$ Subject to: $$3x+4y+s_1=36(1)$$ $$9x+4y+s_2=60(2)$$ $$x,y,s_1,s_2\ge 0$$

Step 2: Write the initial simplex tableau.

The objective function is rewritten as: $$P-50x-60y=0$$ The initial tableau is:

$$\begin{array}{ccccccc}& x& y& s_1& s_2& P& RHS\\ s_1& 3& 4& 1& 0& 0& 36\\ s_2& 9& 4& 0& 1& 0& 60\\ P& -50& -60& 0& 0& 1& 0\end{array}$$

Step 3: Identify the pivot column.

The most negative value in the objective row is $-60$, so the pivot column is $y$.

Step 4: Identify the pivot row.

The ratios of the RHS to the pivot column are: $$\frac{36}{4}=9,\frac{60}{4}=15$$ The minimum ratio is $9$, so the pivot row is the first row.

Step 5: Perform row operations to make the pivot element 1 and eliminate other entries in the pivot column.

Divide the first row by 4: $$\begin{array}{ccccccc}& x& y& s_1& s_2& P& RHS\\ s_1& \frac{3}{4}& 1& \frac{1}{4}& 0& 0& 9\\ s_2& 9& 4& 0& 1& 0& 60\\ P& -50& -60& 0& 0& 1& 0\end{array}$$

Eliminate $y$ from the second row and objective row: $$\text{Row 2}=\text{Row 2}-4\times \text{Row 1}$$ $$\text{Row 3}=\text{Row 3}+60\times \text{Row 1}$$

The updated tableau is:

$$\begin{array}{ccccccc}& x& y& s_1& s_2& P& RHS\\ y& \frac{3}{4}& 1& \frac{1}{4}& 0& 0& 9\\ s_2& \frac{27}{4}& 0& -1& 1& 0& 24\\ P& -25& 0& 15& 0& 1& 540\end{array}$$

Step 6: Check for optimality.

The objective row has no negative values, so the current solution is optimal.

Step 7: Interpret the solution.

The values of the variables are: $$y=9,s_2=24,x=0$$ The maximum value of $P$ is 540.

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Final Answer:

The maximum value of $P$ is 540 when $x=0$ and $y=9$.