differentiation (1 Viewer)

[skyturtle]

heyy~ can sumone please help me with my maths question in advance...> <'

The running cost (cost of fuel) for a certain ship is $3 per hour when the ship is not moving, and this cost increases by an account that is proportional to the cube of its speed, Vkm per hour. If the running cost per hour is $6.75 when the speed is 15km per hour obtain a formula for the running cost per hour at speed V, and calculate the value of V for which the total running cost for a journey of 450km is a minimum.



Thank you ~

 

[kony]

first, the formula for cost, C, per hour is 3 + kV³, for some number k.

therefore, C = 3 + kV³ for one hour.

subsituting C = 6.75 and V for 15,

6.75 = 3 + k15³
4.75 = 3375k

k=19/13500

therefore, C = 3 + 19V³/13500 #


the second part is a bit more complicated.


total cost, c, = (3 + 19V³/13500) * h, where h is the number of hours it takes to complete the job.

but h = 450/V

c = (3 + 19V³/13500)*450/V

dc/dv = (450/v)(19V²/4500) - (3 + 19V³/13500)(450/V²) [product rule]

dc/dv = 1.9V - (3 + 19V³)/30V²

now, for minimum cost, dc/dv = 0

1.9V - (3 + 19V³)/30V² = 0
57V² - 3 - 19V³ = 0
19V³ - 57V² + 3 = 0

well, here, you can do an approximation, using newton's method or trial and error. or use polynomial methods. there doesn't seem to be an integer solution to this though. but, please someone check my working.

alternatively, you should be able to use dc/dv = dc/dh * dh/dv.

 

[ssglain]

6.75 = 3 + k15³
4.75 = 3375k

k=19/13500

therefore, C = 3 + 19V³/13500

3375k = 6.75 - 3 = 3.75
k = 900

That could be why you got such a strange cubic in the end.

 

[kony]

OMG rofl

6.45 - 3 = 4.75 hahaha :rofl:

thanks soph

 

[kony]

double post, but who cares, these forums are like private ones.

C = 3 + V³/900

total cost = (3 + V³/900)h, for number of hours h

since h = 450/V

total cost = (3 + V³/900)*450/V

= 1350/V + V^2/2


differentiating, dc/dv = -1350/V^2 + V

let this = 0.

therefore V = 1350/V^2

V^3 = 1350,
V = 11.05....... km/h