Q. The maximum value of $z = 2x + 5y$ subject to the constraints $2x + 5y \leq 10, x + 2y \geq 1, x - y \leq 4,\, x \geq 0, y \geq 0$, occurs at
Linear Programming
Solution:
We find that the feasible region is on the same side of the line $2x + 5y = 10$ as the origin, on the same side of the line $x - y = 4$ as the origin and on the opposite side of the line $x + 2y = 1$ from the origin. Moreover, the lines meet the coordinate axes at $(5, 0),(0, 2); (1, 0), ( 0, 1/2)$ and $(4, 0)$.
The lines $x - y = 4$ and $2x + 5y = 10$ intersect at $\left( \frac{30}{7} , \frac{2}{7} \right) $
The values of the objective function at the vertices of the pentagon are:
(a) $0 + \frac{5}{2} = \frac{5}{2} $
(b) $2 + 0 = 2 $
(c) $8 + 0 = 8$
(d) $\frac{60}{7} + \frac{10}{7} = 10$
(e) $0 + 10 = 10$
The maximum value 10 occurs at the points $D \left( \frac{30}{7} , \frac{2}{7} \right)$ and $E (0,2),$ Since D and E are adjacent vertices, the objective function has the same maximum value 10 at all the points on the lines DE.
