Minimize: 5x + 10y
Subject to: (Constraints) 2x + y ≥ 10
× + 3y ≥ 15
x ≥ 0 ; y ≥ 0
I'm having trouble determining the feasible region.
I chose point (0,0) and substitute it to the constraints however it does not satisfy the inequality thus FALSE and there's nothing to shade.
Is there a possibility to have a feasible region here. If so, please kindly explain to me. Thank you!
First draw the line 2x+ y= 10. When x= 0, 2(0)+ y= y= 10. (0, 10) is on the line. When y= 0 2x+ 0= 2x= 10 so x= 5. (5, 0) is on the line. Draw the straight line between (0, 10) and (5, 0). Since (0, 0) does not satisfy the inequality the feasible region is on the other side of that line, above and to the right of that line.
Second draw the line x+ 3y= 15. When x= 0, 0+ 3y= 3y= 15 so y= 5. (0, 5) is on the line. When y= 0, x+ 3(0)= x= 15. (15, 0) is on the line. Draw the straight lie between (0, 5) and (15, 0). Since (0, 0) does not satisfy the inequality the inequality the feasible region is on the other side of that line, above and to the right of that line.
Notice that those two lines intersect in the first quadrant. At the point of intersection, the two equations, 2x+ y= 10 and x+ 3y= 15 are both true. From 2x+ y= 10, y= 10- 2x so x+ 3(10- 2x)= x+30- 6x= -5x+ 30= 15. -5x= -15 so x= 3. The y= 10- 2(3)= 4.
The feasible region is above and to the right of those two lines. The basic concept of "Linear Programming" is that a min or max of a linear function, on a convex set, always occurs at a vertex of the convex set. The three vertices of the feasible region here are (0, 10), (3, 4), and (15, 0).
At (0, 10), 5x+ 10y= 5(0)+ 10(10)= 100.
At (3, 4), 5x+ 10y= 5(3)+ 10(40= 55.
At (15, 0). 5X+ 10y= 5(15)+ 10(0)= 75.
The minimum is 55 at (3, 4) and the maximum is 100 at (0, 10).
