The speed of a man carrying an x kg load is given by f(x) = 1/(x+3) m/s. How can the person carry 100 kg across 250 m in the least amount of time.
The speed of the person when he carries x kg is given by `1/(x+3) ` m/s. He has to move a 100 kg load across a distance of 250 m. The problem seems to be one where it has to be determined whether the load should be carried as 100 kg in one go or split up into smaller portions to reduce the time.
If he carries 100 kg, his speed is `1/(103)` m/s. The time taken is `250/(1/103) = 25750` s.
Consider the case where the person decides to carry the load in two portions of 50 kg. Here his speed is `1/53` m/s. The time taken by him is 250*53 = 13250 s. To carry the other portion he would have to return which would take `250/(1/3) = 750` s. The total time taken in this case to carry the entire 100 kg is 13250*2 + 750 = 27250 s.
It is seen that just dividing the load into two increases the total time required to carry the 100 kg across 250 m. Dividing the load into smaller portions would further increase the total time.
To minimize the total time the person should carry the 100 kg load in one go and this takes 25750 s which is the least time required.