We have that Bob is 30 miles from A along the beach and Alice is 2.75 miles from A perpendicular to the beach.

If they meet at point D, X miles from A in the direction of B, then since Bob walks at 3.25 miles/hr, the time he takes to get to D in hours is

T1 = (30-X)/3.25

Since Alice rows at 2.25 miles/hr, the time she takes to reach D in hours is

T2 = sqrt(X^2+2.75^2)/2.25

Plot these on a graph

We see that they should arrive at point D at the same time to minimise the time they take to meet. If one has to wait for the other, then they will take longer to meet.

Therefore we need to solve T1=T2 for X

This gives

`sqrt(x^2+2.75^2)/2.25 = (30-x)/3.25`

`implies` `x^2+2.75^2 = (2.25/3.25)^2(30-x)^2`

`implies` `x^2 +2.75^2 = (2.25/3.25)^2 (900-60x+x^2)`

`implies` `x^2(1-(2.25/3.25)^2) +60(2.25/3.25)^2x +2.75^2-900(2.25/3.25)^2 =0`

Now, letting `a= (1-(2.25/3.25)^2)`, `b = 60(2.25/3.25)^2` and `c=2.75^2-900(2.25/3.25)^2`

solve using the quadratic formula

`x = (-b+-sqrt(b^2-4ac))/(2a)`

` `This gives `x=12.09` miles from A towards B

Check `(30-12.09)/3.25 = 5.51` hours

and `sqrt(12.09^2+2.75^2)/2.25 = 5.51` hours

**They should meet 12.09 miles along the beach from point A, (30-12.09) miles along the beach from point B. It takes 5 and a half hours**.

Bob is at a point B 30 miles from point A and Alice is in a boat 2.75 miles from point A. It is assumed that the lines AC and AB are perpendicular to each other. Alice can row the boat at 2.25 miles per hour and Bob walks at 3.25 miles per hour.

Alice rows at a speed slower than that at which Bob can walk but Bob is further away from point A than Alice. The point chosen to meet should be such that the two can reach it in the same duration of time in spite of the difference in their speed and the fact that they are initially at a different distance from A.

Let the point at which they meet in the least amount of time be X. The distance that Bob has to walk in this case is 30 - X; it can be done in `T = (30 - X)/3.25` hours. The distance that Alice has to row the boat is `sqrt(X^2 + 2.75^2)` . It can be done in `T = sqrt(X^2 + 2.75^2)/2.25` hours.

Equating the two gives `sqrt(X^2 + 2.75^2)/2.25 = (30 - X)/3.25`

=> `X = (13*sqrt(502307)-4320)/532 ~~ 9.19` miles.

**The distance of the landing point of Alice from A should be approximately 9.19 miles.**