#### Linear programming

Q. No. 9,10 and 11 9.  let distance covered with speed 30km/hr= X

let distance covered with speed 40km/hr= Y

Maximum distance will be =X+Y----------------i

Total money to be spend =120

Cost incurred to cover 30 km =2X

Cost incurred to cover 40 km =4Y

2X+4Y=120--------------ii

Total time will be less or equal to 1 hour

Now Distance =Speed *Time

Time =Distance/Time

Time taken with 30km per hour =X/30

Time taken with 40km per hour =Y/40

(X/30)+( Y/40)<=1

4X+3Y<=120------------iii

Take Second equation put X=0 then Y=30      , if we put Y=0 then X=60

X= 0,  Y=30    ( 1st  POINT)

Y=0, x=60        ( 2nd point)

Take third  equation put X=0 then Y=30      , if we put Y=0 then X=60

X= 0,  Y=40    (3rd   POINT)

Y=0, x=30        ( 4th  point)

Solve equation ii and iii

X=12 , y=24  ( 5th point)

Put  the values of points in equation  i

X= 0,  Y=30       Profit=0+30=30

X=60, Y=0         Profit=60+0=60

Y=0 , X =250        Profit=250+0=250

X=0,Y=230           Profit=0+230=230

Y=0, X=276          Profit=276+0=276 (maximum)

Plot all point on graph paper