Need help with applications of calculus question

[preliminary17hsc18]

HI all,
just need some help with this question,

A truck is to travel 1000km at a constant speed of v km/h. When travelling at v km/h, the truck consumes fuel at a rate of (6 + v^2/50) litres per hour. The truck company pays 50c/litre for fuel and pays each of the two drivers $20 hour.
i. Let the total cost of fuel and the wages be C dollars. Show that C = 10v + 43000 x 1/v

Thanks

[Opengangs]

Hey, preliminary17hsc18!

So, let's take it step by step. If we define \( C \) to be the cost of the fuel and the wages, then:
\[ C = \text{cost of fuel} + \text{wages}. \]

To find the cost of fuel, we know that the truck company pays $0.50 per litre for fuel.
We also know that the truck runs at  rate of \( (6 + \frac{v^2}{50})\text{litres} \) per hour.
So, all we need to do is to find the total number of hours it runs for the duration of their journey.

This is where our principle distance-time-speed ratio comes in handy.
\[ \text{If } D = S \cdot T, \\ T = \frac{D}{S}. \]

\[ \text{Thus, it takes the truck: } \\ \frac{1000}{v} \text{ hours to complete their journey.} \]

In finding this, our cost of fuel becomes:
\[ \frac{1000}{v} \cdot \left(6 + \frac{v^2}{50}\right) \cdot 0.5 = \frac{3000}{v} + 10v. \]

Now, the wages is simply the time it takes to complete the journey multiplied by 20 multiplied by 2 (for both drivers).

So:
\[\begin{align*} C &= \frac{3000}{v} + 10v + \frac{1000}{v} \cdot 20 \cdot 2 \\ &= \frac{3000}{v} + 10v + \frac{40000}{v} \\ &= 10v + \frac{43000}{v}. \end{align*}\]

Hopefully, this helps. Let me know if you're stuck with anything