Max Z = 2X + 5Y
S. a
2X – 4Y ≤ 8
-3X + 8Y ≤ 11
X, Y ≥ 0
Accepted Solution
A:
The vertices are $$ \left(0,-2\right);\left(4,0\right);\left(27,11.5\right);\left(0,\frac{11}{8}\right);\left(-\frac{11}{3},0\right) $$
$$ \max\left(z\right)=\left(27,11.5\right)=111.5 $$