Step 1:

It is given that the normal speed of the motorcycle is $50km/hr$

His increased speed is 80km/hr

he has at most of 1hour's time.

Let $x$ and $y$ be the time,which we takes,when the speed is normal

$\therefore 80x+50y\leq 4000$

$\Rightarrow 8x+5y\leq 400$

The cost for the petrol when goes in the normal speed is Rs.2 per kms.

The cost for the petrol when goes in the increased speed is Rs.3 per kms.

He has at most of Rs 120 to spend.

$\therefore 2x+3y\leq 120$

Step 2:

Let us draw the graph for the lines $AB :8x+5y=400$ and $CD=2x+3y=120$

Let us solve these two lines to find the point of intersection

$8x+5y=400$-----(1)

$2x+3y=120$------(2)

Multiply equ(2) by 4

$8x+5y=400$

$8x+12y=480$

_________________

$-7y=-80$

$y=\large\frac{80}{7}$$\Rightarrow x=\large\frac{300}{7}$

Step 3:

Consider the line $AB: 8x+5y=400$

Put $x=0,y=0$ then $0\leq 400$ which is true.

Hence the region $8x+5y\leq 400$ lies below the line AB

Consider the line $CD :2x+3y=120$

Put $x=0,y=0$,then $0\leq 120$,which is true.

Hence the region $2x+3y\leq 400$ lies below the line CD.

The region $O(0,0)$ is also included in the feasible region.

The feasible region is the shaded portion OCPB as shown in the fig.

The corner points of the feasible region are $O(0,0),C(0,40),P(\large\frac{80}{7},\frac{300}{7})$$,B(50,0)$

Step 4:

Now let us find the value of the objective function $Z=x+y$ subjected to constraints $8x+5y\leq 400,2x+3y\leq 120,x,y\leq 0$

At the points(x,y) the value of the objective function subjected to $Z=x+y$

At $O(0,0)$ the value of the objective function $Z=0$

At $C(0,40)$ the value of the objective function $Z=0+40=40$

At $P(\large\frac{300}{7},\frac{80}{7})$ the value of the objective function $Z=\large\frac{300}{7}+\frac{80}{7}=\frac{380}{7}=$$54\large\frac{2}{7}$

At $B(50,0)$ the value of the objective function $Z=50+0=50$

Hence the maximum distance is $54\large\frac{2}{7}$km