math question ?

Cₖ: toal cost per km in £/km
Cₕ: total cost per hr in £/hr
C₁ : cost of fuel per hr
s : speed in km/hr

C₁ ∝ s²
Then, C₁ = ks² where k is a non-zero constant.
When s = 40, C₁ = 25:
25 = k(40)²
k = 1/64
Hence, C₁ = s²/16

Cₕ = C₁ + 100
Cₕ = (s²/16) + 100

Cₖ = Cₕ/s
Cₖ = [(s²/16) + 100]/s
Cₖ = [s² + 1600]/(16s)
Cₖ = {[s² - 2(40)s + 40²] + 2(40s)}/(16s)
Cₖ = [(s - 40)² + 80s]/(16s)
Cₖ = [(s - 40)²/16s] + 5

For any positive value of s, [(s - 40)²/16s] ≥ 0
Hence, Cₖ = [(s - 40)²/16s] + 5 ≥ 5
Cₖ is a minimum when (s - 40)² = 0
Hence, the total cost per km is a minimum when the speed is 40 km/hr.

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OR: Using differentiation

Cₖ = [(s²/16) + 100]/s
Cₖ = (s/16) + (100/s)
Cₖ' = (1/16) - (100/s²) = (s + 40)(s - 40)/(16s²)
Cₖ" = 200/s³

When s = 40:
Cₖ' = 0
Cₖ" = 200/s³ for s > 0
Hence, the total cost per km is a minimum when the speed is 40 km/hr.

Mostly good math in your 1st answer, but 
they made a mistake at the beginning and 
then propagated the error throughout the rest 
of their work.  I think unless 
another answer explains the reasoning.  
k = 1/64

Hence, C₁ = s²/16  (no, it doesn't) 
No, why would be this?  Why did he multiply by 4.
C per km  = (1/64)s^2   
Fuel cost per hour = (1/64)(s^2) £/hour
Fuel cost per hour = (1/64)s^2 / (s km/hr) = (1/64)s
Fixed per km = 100 £/hr/ (s hr/km) = 100/s  

Then,  
Total Cost Per Km = = (1/64)s + 100/s
d (TC Per Km)/ dx =  (1/64)  -100/s^2  
=  (s^2 -100*64)/64s^2
= (s^2 -6400) /64s^2  
so this is zero when the numerator is 0 
s^2 -6400  = 0   
s^2 = 6400 
s = +/- sqrt(6400) = +/- 80 
but negative -80 is not possible 
s = 80 km/hour  
at S = 80 km/hours 
C per km = (1/64)*(80^2) =  100 pounds/km 
Fuel C per  hour  = 100 pounds/km / (80 km/hour) 
Fuel C per hour  =  1.25 Pounds /hour 
Other Cost/hour  = 100/80 = 1.25 Pounds /hour  

Total Cost per Hour = 1.25  + 1.25  = 2.5 Pounds/hour 

so 80 km/hour is the optimum speed  
for best cost/km.