The Point of Maximum Travel. HELP PLEASE !!!!!!!!?

Question

An aircraft is capable of cruising at a steady speed of 300 km per hour in still air. There is enough fuel on board to last 4 hours. On the outward journey, you are helped along by a tail wind of 50km/h which increases your cruising speed relative to the ground to 350km/h. you then realise that on the inward journey you will be flying into a head wind of 50km/h which will reduce your speed to 250km/h. What is the maximum distance you can fly from the airfeild and still be sure of having enough fuel to get back? Investigate these points of maximum distance for different wind speeds.

Answer 1

Formula for the time on the outbound trip (speed = 300 + 50)
t1 = D / 350
Formula for the time on the return trip (speed = 300 - 50)
t2 = D / 250
Formula for total time.
t1 + t2 = 4

So now have three equations... plug the first two into the third and solve.
D / 350 + D / 250 = 4

Common denominator is 1750, so multiply by 1750:
5D + 7D = 4 * 1750
12D = 7000
D = 7000 / 12
D = 583 1/3 miles

For the general case, and a windspeed of x, you have:
4 = D / (300 + x) + D / (300 - x)

The common denominator is 300² - x²:

4 ( 300² - x² ) = D(300 - x) + D(300 + x)
4 ( 300² - x² ) = 300D - Dx + 300D + Dx
4 ( 300² - x² ) = 600D
(300² - x²) / 150 = D

f(x) = (90000 - x²) / 150

As a double-check, put in our values from before of x = 50 km/h
f(50) = ( 90000 - 2500 ) / 150
f(50) = 87500 / 150
f(50) = 583 1/3

Answer 2

Let us designate the speed of aircraft in steady air as V (V=300km/h), and the speed of wind as Vw.
Then the time of flying in outward direction is Tout, the distance in outward direction will be L = (V + Vw)Tout.
The same distance should be overcome in order to come back to the airfield is the same, L. The speed of flying inward will be V - Vw, and it will take time to come back Tin = L/(V - Vw).
The total time of fly (Tout + Tin) should not exceed 4 hours. So, at the maximum available distance Lmax we can write
Tout + Tin = 4.
So, Lmax/(V + Vw) + Lmax/(V - Vw) = 4
(a) First, let us find Lmax for V = 300 km/h, Vw = 50 km/h:
Lmax/350(km/h) + Lmax/250(km/h) = 4(h)
or Lmax (1/350 + 1/250 ) = 4
Lmax (5/1750 + 7/1750) = 4
Lmax ( 12/1750 ) = 4
L max = 4*1750/12 = 1750/3 = 583.33(3) (km)
That is the answer for the first part of the question.
(b) to answer the second, let us write the equation for the Lmax as the function of Vw:
Lmax/(V + Vw) + Lmax/(V - Vw) = 4(h)
Lmax(1/(V + Vw) + 1/(V - Vw)) =4(h) Here L is in km, V and Vw in km/h, and write part of equation in h. Further, we shall remember this.
Lmax[(V-Vw) + (V+Vw)]/[(V-Vw)(V+Vw)] = 4
Lmax(2V)/(V^2 - Vw^2) = 4
Lmax = 4(V^2 - Vw^2)/(2V) = 2(V^2 - Vw^2)/V = 2V - 2Vw^2/V (km)
That is the needed formula.
We can see that the distance is decreases as the speed of wind increases. The maximum distance is in the case when there is no wind at all. And the plane can con fly and then return when the speed of wind equals the speed of plane in steady air.