#### Linear programming

Question 19 Let number of First type of bond= X

Let number of Second type of bond= Y

Therefore X+Y=80 (in thousand) -------------i

Minimum Investment on X => 12 ( in thousand)------ii

Maximum Investment on Y =< 50 (in  thousand) ------iii

Maximum Profit =8%X+10%Y -----------------iv

Take First equation put X=0 then Y=80      , if we put Y=0 then X=80

X= 0,  Y=80    ( 1st  POINT)

Y=0, x=80        ( 2nd point)

Take First equation put X=12 then Y=68      , if we put Y=50 then X=30

X= 12,  Y=68   (3rd   POINT)

Y=50, x=30        ( 4th  point)

From Equation ii we get X=12 ,Y=0 (5th Point)

From Equation iii we get X=0 ,Y=50 (6th Point)

Put the values of points in equation  iv

X= 0,  Y=80       Profit=8%0+10%80=8 (Thousand)  Maximum

X=80, Y=0         Profit=8%80+0=6.4 (Thousand)

Y=68, X =12        Profit=8%68+10%12=6.64(Thousand)

X=30,Y=50           Profit=8%30+10%50=7.4(Thousand)

Y=0, X=12          Profit=8%12+10%0=.96 (Thousand)

Y=50, X=0          Profit=0+8%50=4(Thousand

Plot all point on graph paper