find the maximum value of\np = 6x + 2y\nsubject to the following constraints:\n\\begin{cases}\n4x - y \\geq…

Explanation:

Step1: Analyze the constraints

First, we have the constraints:

• (4x - y \geq 1) can be rewritten as (y \leq 4x - 1)
• (x \geq 0)
• (x \leq 5)
• (y \geq 0)

We need to find the feasible region by graphing these inequalities (or analyzing the vertices). The vertices of the feasible region will be the points where we can check for the maximum of (P = 6x + 2y).

Step2: Find the vertices of the feasible region

1. Intersection of (x = 5), (y = 0), and (4x - y \geq 1): Substitute (x = 5) and (y = 0) into (4x - y): (4(5)-0 = 20\geq1), so this point ((5,0)) is in the feasible region.
2. Intersection of (x = 5) and (y = 4x - 1): Substitute (x = 5) into (y = 4x - 1), we get (y = 4(5)-1=19). So the point is ((5,19))
3. Intersection of (y = 0) and (4x - y = 1): Substitute (y = 0) into (4x - y = 1), we get (4x=1), so (x=\frac{1}{4}). The point is ((\frac{1}{4},0))
4. Intersection of (x = 0) and (y \geq 0) and (4x - y \geq 1): When (x = 0), (4(0)-y\geq1) implies (y\leq - 1), but (y\geq0), so no intersection here.

So the vertices are ((\frac{1}{4},0)), ((5,0)) and ((5,19))

Step3: Evaluate (P = 6x + 2y) at each vertex

1. At ((\frac{1}{4},0)): (P=6(\frac{1}{4})+2(0)=\frac{6}{4}=\frac{3}{2} = 1.5)
2. At ((5,0)): (P=6(5)+2(0)=30)
3. At ((5,19)): (P=6(5)+2(19)=30 + 38=68)

Answer:

(68)