Question

1. maximize the equation $p = 3x + 2y$ using the constraints shown.\\\

$$\begin{cases}x + 2y \\geq 8\\\\x \\geq -2\\\\y \\leq 5\\\\x \\leq 14\\end{cases}$$

\\\\(\boldsymbol{p = 52}\\)\\(\boldsymbol{p = 58}\\)\\(\boldsymbol{p = 36}\\)\\(\boldsymbol{p = 4}\\)

Explanation:

Step1: Identify Feasible Region Vertices

First, find intersection points of constraints:

• Intersection of \(x = 14\) and \(y = 5\): \((14, 5)\)
• Intersection of \(x = 14\) and \(x + 2y = 8\): Solve \(14 + 2y = 8\) → \(2y = -6\) → \(y = -3\) (but \(y \leq 5\) and \(x + 2y \geq 8\), check if valid. However, \(y = -3\) is below, but \(x = 14\), \(y = 5\) is better for maximization as \(P = 3x + 2y\) increases with \(x,y\) (coefficients positive).
• Intersection of \(y = 5\) and \(x + 2y = 8\): \(x + 10 = 8\) → \(x = -2\), so \((-2, 5)\)
• Intersection of \(x = -2\) and \(x + 2y = 8\): \(-2 + 2y = 8\) → \(2y = 10\) → \(y = 5\) (same as above)
• Intersection of \(x = -2\) and \(y = 5\): \((-2, 5)\) (already found)
• Intersection of \(x = 14\) and \(y = 5\) is key for max (since \(x\) and \(y\) are as large as constraints allow, given \(3x + 2y\) has positive coefficients).

Step2: Calculate \(P\) at Vertices

• At \((14, 5)\): \(P = 3(14) + 2(5) = 42 + 10 = 52\)
• At \((-2, 5)\): \(P = 3(-2) + 2(5) = -6 + 10 = 4\)
• At intersection of \(x = 14\) and \(x + 2y = 8\) (invalid \(y\) as above, \(P\) would be lower)
• Check other points, but \((14,5)\) gives highest \(P\) among valid vertices.

Answer:

\(P = 52\) (corresponding to the option \(P = 52\))