Question

The maximum value of Z = 10x + 16y, subject to constraints $ x \geq 0, \quad y \geq 0, \quad x + y \leq 12, \quad 2x + y \leq 20 $

• A. 144
• B. 192
• C. 120
• D. 240

Hint:

In linear programming, always plot the feasible region and find the vertices of the region to maximize or minimize the objective function.

Answer 1

The Correct Option is B

To determine the maximum value of the objective function \( Z = 10x + 16y \) at the given vertices of the feasible region:

1. Evaluate at Origin (0,0):
\[ Z = 10(0) + 16(0) = \boxed{0} \]

2. Evaluate at Point A (10,0):
\[ Z = 10(10) + 16(0) = \boxed{100} \]

3. Evaluate at Point B (8,4):
\[ Z = 10(8) + 16(4) = 80 + 64 = \boxed{144} \]

4. Evaluate at Point C (0,12):
\[ Z = 10(0) + 16(12) = \boxed{192} \]

5. Determine Maximum Value:
Comparing all calculated values:
\[ 0 < 100 < 144 < 192 \]
The maximum value occurs at point C (0,12).

Final Answer:
The maximum value of \( Z \) is \(\boxed{192}\).

Answer 2

Finding the Vertices:

• O: (0, 0) (Intersection of \(x=0\) and \(y=0\))
• A: (10, 0) (Intersection of \(2x + y = 20\) and \(y=0\))
• B: (8, 4) (Intersection of \(x+y=12\) and \(2x+y=20\). Subtracting the first equation from the second gives \(x=8\), and substituting back into \(x+y=12\) yields \(y=4\)).
• C: (0, 12) (Intersection of \(x+y=12\) and \(x=0\))

Evaluating Z at each vertex:

• O: \(Z(0,0) = 10(0) + 16(0) = 0\)
• A: \(Z(10,0) = 10(10) + 16(0) = 100\)
• B: \(Z(8,4) = 10(8) + 16(4) = 80 + 64 = 144\)
• C: \(Z(0,12) = 10(0) + 16(12) = 192\)

The maximum value of \(Z\) is 192, which occurs at point C (0, 12).