First, find the vertices of the shaded region.
x≥0, y≥0 represents the first quadrant.
x + y ≤ 11 has intercepts at (11,0) and (0, 11)
2y ≥ x has intercepts at (0, 0)
x + y ≤ 11 and 2y ≥ x intercept at [tex](\frac{22}{3}, \frac{22}{6} )[/tex]
The vertices within the shaded region are (0, 11), (0, 0), and [tex](\frac{22}{3}, \frac{22}{6} )[/tex].
Next, evaluate the function P = 2x + y at those intercepts.
P = 2x + y at (0, 11) ⇒ P = 2(0) + (11) ⇒ P = 0 + 11 ⇒ P = 11
P = 2x + y at (0, 0) ⇒ P = 2(0) + (0) ⇒ P = 0 + 0 ⇒ P = 0
P = 2x + y at [tex](\frac{22}{3}, \frac{22}{6} )[/tex] ⇒ P = 2[tex](\frac{22}{3})[/tex] + [tex](\frac{22}{6} )[/tex] ⇒ P = [tex]\frac{44}{3}[/tex] + [tex]\frac{22}{6} [/tex] ⇒ P = [tex]\frac{110}{6}[/tex] ⇒ P = [tex]\frac{55}{3} [/tex] = 19[tex]\frac{1}{3}[/tex]
The maximum is found at [tex](\frac{22}{3}, \frac{22}{6} )[/tex]
Answer: maximum is 19[tex]\frac{1}{3}[/tex]
