# The cost of fuel is running an engine is proportional to the square of the speed in km/h is ₹48 per hour when the speed is 16 kn/h.Other costs amount to ₹300 per hour.Find the most economical speed?

Jun 30, 2018

$40 \text{ kph}$

#### Explanation:

Don't know that currency symbol so the euro, €, is used here as a proxy currency.

"The cost of fuel is running an engine is proportional to the square of the speed..."

• $\implies C = k {v}^{2}$

And:

•  48 \ (€)/(hr) harr k ( 16 (km)/(hr))^2 implies k = 3/16

"Other costs amount to €300 per hour":

• $\implies C = \frac{3}{16} {v}^{2} + 300$

But that equation is useless because it measures the cost per hour.

This silly example makes the point. If you are going to cost-optimise a use of the car for 100hours, you may as well just sit in it and pay only the other costs. You will go nowhere but you are not paying for fuel.

For this optimisation to make any physical sense, it should focus on true fuel efficiency, ie the all-in cost of travelling a given distance.

Note:

(€)/(hr) = color(red)((€)/(km))(km)/(hr) implies (€)/(km) = ((€)/(hr))/((km)/(hr)) = ((€)/(hr))/v

So with $\overline{C}$ denoting the cost per km travelled:

• $\overline{C} = \frac{C}{v}$

• $\overline{C} = \frac{3}{16} v + \frac{300}{v}$

graph{y = 3/16 x + 300/x [-5, 200, -5, 50]}

Optimising by differentiating wrt $v$:

• $\frac{d \overline{C}}{\mathrm{dv}} = \frac{3}{16} - \frac{300}{v} ^ 2 = 0$

implies v_("opt") = 4 0 " kph", qquad "and " bar C_("opt") = 15 €/(km)