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

Sketch the graph for

x = 0
y =  0
2y = x
x + y = 11

The region that qualifies all the inequalities are labeled R in the diagram below 

The maximum coordinate for P is the intersection of 2y = x and x + y = 11

Let 2y = x be EQ(1) and x + y = 11 is EQ(2)

Substituting EQ(1) into EQ(2)

2y + y = 11
3y = 11
y = ¹¹/₃ ⇒ Substitute this back into either EQ(1) or EQ(2)

2(¹¹/₃) = x
x = ²²/₃

The value of P when x = ²²/₃ and y = ¹¹/₃ is given

P = 3(²²/₃) + 2(¹¹/₃) = 22 + ²²/₃ = 29¹/₃

Option D