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 $P(\large\frac{30}{7},\frac{30}{7})$