Graph the equations or solve the system of equations to find the vertices within the shaded region.
x≥0, y≥0 represents the first quadrant.
x + 2y ≤ 14 ⇒ has intercepts (14, 0) and (0, 7).
4x - y ≤ 20 ; has intercepts (5, 0) and (-20, 0). Disregard (-20, 0) since it is not in Quadrant 1.
x + 2y ≤ 14 and 4x - y ≤ 20 intercept at (6,4).
Now evaluate the value (C) at the intercepts (0, 7), (5, 0), and (6,4). Note: (14, 0) has been disregarded because it is outside of the shaded region.
C = 6x - 2y at (0, 7) ⇒ C = 6(0) - 2(7) ⇒ C = 0 - 14 ⇒ C = -14
C = 6x - 2y at (5, 0) ⇒ C = 6(5) - 2(0) ⇒ C = 30 - 0 ⇒ C = 30
C = 6x - 2y at (14, 0) ⇒ C = 6(6) - 2(4) ⇒ C = 36 - 8 ⇒ C = 28
The maximum value (C) equals 30
Answer: maximum value is 30
