Question

When travelling $x{\text{ km/hour,}}$ a truck burns diesel at the rate of $\dfrac{1}{{300}}\left( {\dfrac{{900}}{x} + x} \right)$ litres per km. If the diesel oil costs $40p.$ per liter and driver is paid Rs. $1.50$ per hour, find the steady speed that will minimize the total cost of the trip of $500{\text{ km}}{\text{.}}$
(A) $15{\text{ km}}{\text{.p}}{\text{.h}}$
(B) ${\text{27 km}}{\text{.p}}{\text{.h}}$
(C) ${\text{42 km}}{\text{.p}}{\text{.h}}$
(D) $45{\text{ km}}{\text{.p}}{\text{.h}}$

Answer 1

First of all for the given trip we will find the diesel cost and then the total cost. To get the minimum cost of the steady speed we will use the differentiation with reference to “x” and will simplify for the required value.

Complete Step By Step Answer:
For the trip of $500{\text{ Km}}$
Diesel Cost can be given by $= \dfrac{1}{{300}}\left( {\dfrac{{900}}{x} + x} \right)\dfrac{{litre}}{{km}} \times 0.4\dfrac{{Rs.}}{{litre}} \times 500km$
Simplify the above equation. Common multiples from the numerator and the denominator cancel each other. Also, the common unit from the numerator and the denominator cancel each other.
Diesel Cost $= \dfrac{2}{3}\left( {\dfrac{{900}}{x} + x} \right)Rs.$ ….. (A)
Now, the Driver Cost $= \dfrac{{500}}{x} \times 1.50{\text{ Rs}}{\text{.}}$
Simplify the above expression-
the Driver Cost $= \dfrac{{750}}{x}{\text{ Rs}}{\text{.}}$ …… (B)
By using equations (A) and (B)
The Total cost $= \dfrac{2}{3}\left( {\dfrac{{900}}{x} + x} \right) + \dfrac{{750}}{x}$ Rs.
Simplify the above expression –
$C = \dfrac{2}{3}x + \dfrac{{1350}}{x}$
For the minimization of the cost, differentiate the above expression with respect to “x”
And placing $\dfrac{{dC}}{{dx}} = 0$
$\dfrac{{dC}}{{dx}} = - \dfrac{{1350}}{{{x^2}}} + \dfrac{2}{3}$
$- \dfrac{{1350}}{{{x^2}}} + \dfrac{2}{3} = 0$
Make the required term “x” the subject –
$\dfrac{{1350}}{{{x^2}}} = \dfrac{2}{3}$
Cross multiply the above equation, where the denominator of one side is multiplied with the numerator of the opposite side.
${x^2} = \dfrac{{1350 \times 3}}{2}$
Take square-root on both the sides of the equation.
$\sqrt {{x^2}} = \sqrt {\dfrac{{1350 \times 3}}{2}}$
Simplify –
$\sqrt {{x^2}} = \sqrt {2025}$
$\Rightarrow x = 45\,{\text{km/hour}}$
From the given multiple choices, the option D is the correct answer.

Note :
Always remember that square and square root cancel each other. Be good in basic mathematical simplification. Also, remember the common factors from the numerator and the denominator cancel each other. Read the question twice and frame the mathematical expressions properly. Be careful about the sign convention and remember when you move any term from one side to another then the sign of the term also changes.