Question

find the minimum value of
$c = 3x + 10y$
subject to the following constraints:
\

$$\begin{cases} 2x + 4y \\geq 20 \\\\ 2x + 2y \\leq 16 \\\\ x \\geq 2 \\\\ y \\geq 3 \\end{cases}$$

$c = ?$

Explanation:

Step1: Simplify Constraints

First, simplify the inequalities. For \(2x + 4y \geq 20\), divide by 2: \(x + 2y \geq 10\). For \(2x + 2y \leq 16\), divide by 2: \(x + y \leq 8\). We also have \(x \geq 2\), \(y \geq 3\).

Step2: Find Intersection Points

1. Intersection of \(x + 2y = 10\) and \(x + y = 8\):

Subtract the second equation from the first: \((x + 2y)-(x + y)=10 - 8\) → \(y = 2\). But \(y \geq 3\), so this point is invalid.

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

Substitute \(y = 3\) into \(x + 2y = 10\): \(x + 6 = 10\) → \(x = 4\). So point \((4, 3)\).

1. Intersection of \(x + y = 8\) and \(y = 3\):

Substitute \(y = 3\) into \(x + y = 8\): \(x + 3 = 8\) → \(x = 5\). Check \(x + 2y = 5 + 6 = 11 \geq 10\), valid. Point \((5, 3)\).

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

Substitute \(x = 2\) into \(x + 2y = 10\): \(2 + 2y = 10\) → \(y = 4\). Check \(x + y = 6 \leq 8\), valid. Point \((2, 4)\).

1. Intersection of \(x + y = 8\) and \(x = 2\):

Substitute \(x = 2\) into \(x + y = 8\): \(y = 6\). Check \(x + 2y = 2 + 12 = 14 \geq 10\), valid. Point \((2, 6)\). But \(y = 6\), check \(x + 2y = 14 \geq 10\), \(x + y = 8 \leq 8\), \(x \geq 2\), \(y \geq 3\).

Step3: Evaluate C at Valid Points

• At \((4, 3)\): \(C = 3(4)+10(3)=12 + 30 = 42\)
• At \((5, 3)\): \(C = 3(5)+10(3)=15 + 30 = 45\)
• At \((2, 4)\): \(C = 3(2)+10(4)=6 + 40 = 46\)
• At \((2, 6)\): \(C = 3(2)+10(6)=6 + 60 = 66\)

Answer:

42