Let r be the number of minutes of radio and t the number of minutes on TV.
r>2t; r<400; 15r+300t<10000 represent the constraints. In terms of effectiveness TV is 25 times more efficient than radio. Using t as the vertical axis and r as the horizontal axis, the lines t=(1/2)r, 15r+300t=10000, r=400 can be plotted. The lines t=(1/2)r and 15r+300t=10000 intersect when r<400. Substituting t=(1/2)r in the second equation we can see that 165r=10000, so r=2000/33, making t=1000/33 at the intersection. The area between the intersection of the two lines, the r axis, and the line r=400 represents where all the constraints are met.
Assuming r and t to be in whole minutes, we need to find the integers closest to the intersection point within the defined area. 2000/33=60 and 1000/33=30 so the advertising cost is for 60 mins of radio and 30 mins of TV:
Cost=15*60+300*30=900+9000=9900, leaving R100 under budget. But R100 will buy a further 6 minutes of radio advertising so r=66, t=30. 66 minutes is more than twice 30 so the constraint is still valid. 30 minutes of TV advertising has an efficiency equivalent of 25*30=750 radio minutes. Add 66 minutes to this and we have 816 minutes, so the combination is worth 816 minutes of radio time.