Question

The difference between the maximum value and minimum value of the objective function \(z = 3x+5y\) of a linear programming problem subject to constraints \(5x+10y\leq 50\), \(x+y\geq 1\), \(y\leq 4\) and \(x\geq 0,y\geq 0\) is \(3λ\). Then the value of \(λ\) is

• A. \(3\).
• B. \(6\).
• C. \(9\).
• D. \(27\).

Hint:

Find the corner points of the feasible region and compare z at each one.

Answer 1

The Correct Option is C

Step 1: List the corners:
Intersect the bounding lines in pairs and keep only points that satisfy every constraint: $(1,0)$, $(10,0)$, $(2,4)$, $(0,4)$, $(0,1)$.

Step 2: Compare z values:
Make a table: $z=3,\,30,\,26,\,20,\,5$ in that order. The largest is 30 at $(10,0)$ and the smallest is 3 at $(1,0)$.

Step 3: Use the given relation:
Difference $=30-3=27=3\lambda$.

Step 4: Solve:
$\lambda=\frac{27}{3}=9$.

Final Answer:
The value of $\lambda$ is 9 (option C). \[ \boxed{9} \]