Linear Programming Math Example 2

Follow the full solution, then compare it with the other examples linked below.

hard

Minimize

z = 2x + 5y

subject to

x + 2y \geq 6

x + y \geq 4

x \geq 0

y \geq 0

1
Step 1: Find corner points. Intersection of $x + 2y = 6$ and $x + y = 4$ : subtract to get $y = 2$ , $x = 2$ . Point: $(2, 2)$ .
2
Step 2: Other vertices: $(6, 0)$ from $x + 2y = 6$ , $y = 0$ ; $(0, 4)$ from $x + y = 4$ , $x = 0$ .
3
Step 3: Evaluate: $z(2,2) = 14$ , $z(6,0) = 12$ , $z(0,4) = 20$ .
4
Step 4: Minimum is $z = 12$ at $(6, 0)$ .

Answer

z = 12

(6, 0)

For minimization with

\geq

constraints, the feasible region is unbounded on one side. Still evaluate all corner points — the minimum (if finite) occurs at a vertex.

About Linear Programming

Linear programming optimizes a linear objective subject to linear inequality or equality constraints.

Maximize [formula] subject to [formula], [formula], [formula].

A feasible region has vertices at [formula], [formula], [formula], [formula]. Maximize [formula].

A company makes chairs ([formula]40[formula][formula] profit). Each chair takes 2 hours, each table