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 $3\lambda$. Then the value of $\lambda$ is

The given constraints are:
$5x + 10y \leq 50 \Rightarrow x + 2y \leq 10$
$x + y \geq 1$
$y \leq 4$
$x \geq 0, y \geq 0$

The feasible region is bounded by the lines $x + 2y = 10$, $x + y = 1$, $y = 4$, $x = 0$, and $y = 0$. Let us find the corner points of this feasible region:
Intersection of $x = 0$ and $x + y = 1 \Rightarrow (0, 1)$
Intersection of $x = 0$ and $y = 4 \Rightarrow (0, 4)$
Intersection of $y = 4$ and $x + 2y = 10 \Rightarrow x + 8 = 10 \Rightarrow x = 2 \Rightarrow (2, 4)$
Intersection of $y = 0$ and $x + 2y = 10 \Rightarrow x = 10 \Rightarrow (10, 0)$
Intersection of $y = 0$ and $x + y = 1 \Rightarrow x = 1 \Rightarrow (1, 0)$

Now, we evaluate the objective function $z = 3x + 5y$ at each of these corner points:
At $(0, 1)$, $z = 3(0) + 5(1) = 5$
At $(0, 4)$, $z = 3(0) + 5(4) = 20$
At $(2, 4)$, $z = 3(2) + 5(4) = 26$
At $(10, 0)$, $z = 3(10) + 5(0) = 30$
At $(1, 0)$, $z = 3(1) + 5(0) = 3$

From the above values, the maximum value of $z$ is $30$ and the minimum value of $z$ is $3$.

The difference between the maximum and minimum values is:
$30 - 3 = 27$

Given that this difference is equal to $3\lambda$, we have:
$3\lambda = 27 \Rightarrow \lambda = 9$

Answer: $9$.