Edward C. answered • 03/19/15

Tutor

5.0
(433)
Caltech Grad for math tutoring: Algebra through Calculus

Y = 1 is a horizontal line that goes thru the points (0,1) and (2,1)

X = 2 is a vertical line that goes thru the points (2,0) and (2,1)

X + 2Y = 6 is a line slanted down to the right that goes thru the points (0,3) and (6,0)

If you shade in the sides of each line where the inequalities are true you should find that the common area of intersection is a triangle. The vertices of the triangle are at the 3 points of intersection of the 3 lines

Y=1 and X=2 intersect at the vertex (2,1)

Y=1 and X+2Y=6 intersect at the vertex (4,1)

X=2 and X+2Y=6 intersect at the vertex (2,2)

The extrema (maximum and minimum) of the objective function will occur at the vertices, so evaluate C at each vertex

C(2,1) = 3*2 + 4*1 = 6 + 4 = 10

C(4,1) = 3*4 + 4*1 = 12 + 4 = 16

C(2,2) = 3*2 + 4*2 = 6 + 8 = 14

The maximum value of C = 16 and occurs at (4,1)

The minimum value of C = 10 and occurs at (2,1)