Maximize ` z = 10 x + 25 y `, subject to ` x le 3, y le 3, x + y le 5, x ge 0 , y ge 0 `.

Explanation

First we draw the lines AB, CD and EF whose equations are ` x = 13 , y = 3 and x + y = 5 ` respectively.


The feasible region is OAPQDO which is shaped in the figure.
The vertices of feasible reagion are ` O ( 0, 0 ), A ( 3, 0 ) , P, Q and D ( 0 , 3 )`.
P is the point of intersection of the lines ` x + y = 5 and x = 3 `
Substituting ` x = 3 ` in ` x + y = 5 `, we get
` 3 + y = 5 `
` therefore y =2 `
` therefore P-= (3, 2 ) `
Q is the point of intersection of the lines ` x + y = 5 and y = 3 `
Substituting ` y = 3 ` in ` x + y = 5`, we get,
` x + 3 = 5" " therefore x = 2 `
` therefore Q -= (2, 3 ) `
The values of the objective function ` z = 10 x + 25 y ` at these vertices are
` z (O) = 10 ( 0 ) + 25 ( 0 ) = 0 `
`z(A) = 10 (3) +25(0) = 30 `
` z (P) = 10 ( 3) + 25 ( 2) = 30 + 50 = 80`
` z ( Q) = 10 ( 2 ) + 25 ( 3 ) = 20 + 75 =95`
` z ( D) = 10 (0) + 25 ( 3 ) = 75 `
` therefore ` z has maximum value 95, when ` x = 2 and y = 3`.