Answer:
P = 34 is the maximum value
Step-by-step explanation:
Sketch
x + y = 6
with x- intercept = (6, 0) and y- intercept = (0, 6)
2x + 3y = 16
with x- intercept = (8, 0) and y- intercept = (0, [tex]\frac{16}{3}[/tex]
Solve x + y = 6 and 2x + 3y = 16 to find intersection at (2, 4)
The feasible region has vertices at
(0, [tex]\frac{16}{3}[/tex]), (0, 0), 6, 0) and (2, 4)
Evaluate the objective function P = 5x + 6y at each vertex
(0, 0) can be rejected as it will not give a maximum
(0, [tex]\frac{16}{3}[/tex]) → 0 + 32 = 32
(6, 0) → P = 5(6) + 0 = 30
(2, 4) → P = (5(2) + 6(4) = 10 + 24 = 34 ← maximum value
Thus maximum value is P = 34 when x = 2 and y = 4
