By graphing the system of constraints find the values of x and y that maximize the objective function, find the maximum value. x+y< =11 2y> =x x> =0 y> =0 maximum for p=2x+y a. p=10 1/3 b. p=7 1/3 c. p=21 d. p=29 1/3

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

The vertices within the shaded region are (0, 11), (0, 0), and .

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   ⇒  P = 2 +  ⇒   P = +  ⇒   P =   ⇒   P =   = 19

The maximum is found at

maximum is 19

∠cbe and ∠deb are same-side interior angles.

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