Question

1. maximize the equation $p = 3x + y$ using the constraints shown.\

$$\begin{cases}x \\geq 0 \\\\ x \\leq 4 \\\\ y \\geq -1 \\\\ y \\leq 5\\end{cases}$$

$\LXI0 $p = 17$\LXI1 $p = 3$\LXI2 $p = -1$\LXI3 $p = 5$

Explanation:

Step1: Identify Objective Function

We need to maximize \( P = 3x + y \) with constraints \( x \geq 0 \), \( x \leq 4 \), \( y \geq -1 \), \( y \leq 5 \).

Step2: Determine Critical Points

The feasible region is a rectangle with vertices at \((0, -1)\), \((0, 5)\), \((4, -1)\), \((4, 5)\). Evaluate \( P \) at each vertex:

• At \((0, -1)\): \( P = 3(0) + (-1) = -1 \)
• At \((0, 5)\): \( P = 3(0) + 5 = 5 \)
• At \((4, -1)\): \( P = 3(4) + (-1) = 11 \) (Wait, original options have 17? Wait, maybe I misread. Wait, no, wait the constraints—wait, maybe I made a mistake. Wait, wait, the problem's options include \( P = 17 \). Wait, maybe the constraints are different? Wait, no, the given constraints are \( x \geq 0 \), \( x \leq 4 \), \( y \geq -1 \), \( y \leq 5 \). Wait, but \( 3(4) + 5 = 17 \)! Oh right! At \((4, 5)\): \( P = 3(4) + 5 = 12 + 5 = 17 \). Ah, I missed that vertex. So:
• At \((4, 5)\): \( P = 3(4) + 5 = 17 \)

Step3: Compare Values

Among \( -1 \), \( 5 \), \( 11 \) (wait no, \((4,5)\) is \( 17 \)), the maximum is \( 17 \).

Answer:

\( P = 17 \) (the option with \( P = 17 \))