Let x1 represents the number of A
Let x2 represents the number of B
Let x3 represents the number of C
The linear program is
Maximize:z=2x1+5x2+8x3Subjectto:6x1+8x2+4x3≤902x1+x2+4x3≤405x1+3x2+2x3≤60
with all variables non-negative.
Next we compute the optimal solution, to do this we put the linear program in standard form:
Maximize:z=2x1+5x2+8x3+0x4+0x5+0x6Subjectto:6x1+8x2+4x3+x4≤962x1+x2+4x3+x5≤405x1+3x2+2x3+x6≤60
with all variables non-negative.
Subsequently, we obtain our Tableau 1
x4(0)x5(0)x6(0)x12625−5x25813−8x384420x401000x500100x6000109640600
Next we locate the most negative number in the bottom row(-8). Then we compute ratios of positive numbers in our work column. Therefore our pivot element is 4. Using row reduction method, we reduce our pivot element to 1 and every other elements in the work column to 0, thereby generating Tableau 2.
x3(8)x5(0)x6(0)x13042142x24074125−3x3350100x401000x50−1412−12x60001056104080
We notice there is still a negative element in the last row therefore repeating the process above we obtain Tableau 3.
x3(8)x2(5)x6(0)x1307475718726x2401000x3350100x409128−118−573x507−1727−6711x6000108820104
Since there are no negative values in our last row. The solution is optimal. The optimal solution is x1=0,x2=5,x3=8,x4=0,x5=0,x6=0.
Placid should produce 5 of product B and 8 of Product B
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