Question

find the maximum value of
p = x + 4y
subject to the following constraints:
\

$$\begin{cases} 2x + 3y \\geq 24 \\\\ x \\geq 0 \\\\ x \\leq 12 \\\\ y \\leq 10 \\end{cases}$$

p = ?

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 \((0, 8)\).

1. 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 \((12, 0)\).

1. Intersection of \(x = 12\) and \(y = 10\):

The point is \((12, 10)\).

1. Intersection of \(y = 10\) and \(2x + 3y = 24\):

Substitute \(y = 10\) into \(2x + 3y = 24\), we get \(2x + 3\times10 = 24\), \(2x = 24 - 30 = -6\), \(x = -3\) (but \(x\geq0\), so this point is not in the feasible region).

1. Intersection of \(x = 0\) and \(y = 10\):

Check if \(2\times0 + 3\times10 \geq 24\) ( \(30\geq24\), yes). The point is \((0, 10)\).

So the feasible region vertices are \((0, 8)\), \((12, 0)\), \((12, 10)\), \((0, 10)\).

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

1. At \((0, 8)\):

\(P = 0 + 4\times8 = 32\)

1. At \((12, 0)\):

\(P = 12 + 4\times0 = 12\)

1. At \((12, 10)\):

\(P = 12 + 4\times10 = 12 + 40 = 52\)

1. At \((0, 10)\):

\(P = 0 + 4\times10 = 40\)

Answer:

52