Question

For the linear programming problem, \(x+2y\leq 10, 3x+y\leq 12, x,y\geq 0\), the maximum value of \(z = 5x+10y\) occurs at every point on the line segment joining the points..

• A. \((0,0)\) and \((4,0)\)
• B. \((0,0)\) and \((0,5)\)
• C. \((4,0)\) and \((\frac{14}{5},\frac{18}{5})\)
• D. \((0,5)\) and \((\frac{14}{5},\frac{18}{5})\)

Hint:

The objective is parallel to the constraint \(x+2y\le10\), so the maximum occurs along an edge.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
\(z=5x+10y=5(x+2y)\). The constraint \(x+2y\le10\) then limits \(z\le50\).

Step 2: Key Formula or Approach
Find the corner points of the feasible region.

Step 3: Detailed Explanation
Corners: \((0,0)\), \((4,0)\), \((0,5)\) and the intersection of \(x+2y=10\), \(3x+y=12\): \(x=\dfrac{14}{5}\), \(y=\dfrac{18}{5}\).
\(z(0,0)=0\), \(z(4,0)=20\), \(z(0,5)=50\), \(z\left(\tfrac{14}{5},\tfrac{18}{5}\right)=14+36=50\).
The maximum 50 occurs at both \((0,5)\) and \(\left(\tfrac{14}{5},\tfrac{18}{5}\right)\), so it holds at every point on the segment joining them.

Final Answer:
The maximum occurs along the segment joining \((0,5)\) and \((\frac{14}{5},\frac{18}{5})\), option (D). \[ \boxed{(0,5)\ \text{and}\ \left(\dfrac{14}{5},\dfrac{18}{5}\right)\ \text{(D)}} \]