find the maximum value of\np = x + 4y\nsubject to the following constraints:\n\\begin{cases}\n2x + 3y \\geq…

find the maximum value of\np = x + 4y\nsubject to the following constraints:\n\\begin{cases}\n2x + 3y \\geq 24\n\\\\\nx \\geq 0\n\\\\\nx \\leq 12\n\\\\\ny \\leq 10\n\\end{cases}\np = ?

find the maximum value of\np = x + 4y\nsubject to the following constraints:\n\\begin{cases}\n2x + 3y \\geq 24\n\\\\\nx \\geq 0\n\\\\\nx \\leq 12\n\\\\\ny \\leq 10\n\\end{cases}\np = ?

Answer

Explanation:

Step1: Identify the feasible region vertices

First, we find the intersection points of the constraint lines.

  1. Intersection of (2x + 3y = 24) and (x = 0): Substitute (x = 0) into (2x + 3y = 24), we get (3y = 24), so (y = 8). The point is ((0, 8)).
  2. Intersection of (2x + 3y = 24) and (x = 12): Substitute (x = 12) into (2x + 3y = 24), we have (2\times12 + 3y = 24), (24 + 3y = 24), so (y = 0). The point is ((12, 0)).
  3. Intersection of (x = 12) and (y = 10): The point is ((12, 10)).
  4. Intersection of (y = 10) and (2x + 3y = 24): Substitute (y = 10) into (2x + 3y = 24), we get (2x + 3\times10 = 24), (2x + 30 = 24), (2x = -6), (x = -3) (but (x\geq0), so this intersection is not in the feasible region).
  5. Intersection of (x = 0) and (y = 10): Check if (2\times0 + 3\times10\geq24), (30\geq24) (true). The point is ((0, 10)).

Now we have the vertices of the feasible region: ((0, 8)), ((12, 0)), ((12, 10)), ((0, 10)).

Step2: Evaluate (P = x + 4y) at each vertex

  • At ((0, 8)): (P = 0 + 4\times8 = 32)
  • At ((12, 0)): (P = 12 + 4\times0 = 12)
  • At ((12, 10)): (P = 12 + 4\times10 = 52)
  • At ((0, 10)): (P = 0 + 4\times10 = 40)

Answer:

52