Step 1:
The given problem can be written as follows :
Large van $\Rightarrow$ packages :200,cost :Rs.400
Small van $\Rightarrow$ packages :80,cost :Rs.200
maximum package $\Rightarrow$ packages :1200,cost :Rs.3000
The mathematical formulation for the above problem
The objective function to be minimized is $Z=400x+200y$
Let $x$ be the packages carried by large van and $y$ be the packages carried by small van.
Then $200x+80y\geq 1200\Rightarrow 5x+2y\geq 30$
The total cost should not be exceed Rs.3000
Hence $400x+200y\geq 3000$
$2x+y\geq 15$
Since the large vans cannot exceed the small van.
$x\geq y$
Hence the objective function is $Z=400x+200y$ subject to the constraints $5x+2y\geq 30,2x+y\geq 15$ and $x\geq y$ and $x,y\geq 0$
Step 2:
The lines AB :$5x+2y=30,AC :2x+y=15$ and $CQ :x=y$ can be drawn on the graph to find the feasible region.
Consider the line $AB :5x+2y=30$
Put $x=0,y=0$ then $0\geq 30$ is not true.
Hence the region $5x+2y\geq 30$ lies above the line AB.
Consider the line $AC :2x+y=15$
Put $x=0,y=0$ then $0\geq 15$ is not true.
Hence the region $2x+y\geq 15$ lies above the line AC.
Consider the line $CQ :x-y=0$
Put $x=0,y=0$
The region $x\geq y$ lies above CQ.
Clearly the origin $O(0,0)$ is not included.
The feasible region is APQ is the shaded region shown.
The corner points of the feasible region are $A(0,15),P(\large\frac{30}{7},\frac{30}{7})$ and $Q(5,5)$
Step 3:
The value of the objective function can be found as follows :
At the points $(x,y)$ the value of the objective function subjected to $Z=400x+200y$
At $O(0,15)$ the value of the objective function $Z=400\times 0+200\times 15=3000$
At $P(\large\frac{30}{7},\frac{30}{7})$ the value of the objective function $Z=400\times \large\frac{30}{7}$$+200\times \large\frac{30}{7}$$=1714.2+857.1=2571.3$
At $Q(5,5)$ the value of the objective function $Z=400\times 5+200\times 5=3000$
Hence the maximum value is $2571.3$ at $P(\large\frac{30}{7},\frac{30}{7})$