Fortuner Fortunes

I think you should provide a better description for how we should write the answer. For instance, if the answer is $x$ , enter the answer as $\lfloor x \rfloor$ .

Anyway, the total distance that was covered was $800 \times 2 = 1600$ km. The time required for this journey, if we traveled at $X$ km/hr is $\dfrac{1600}{X}$ hours. The driver charges Rs 125 per hour, so the total charges will be $125 \times \dfrac{1600}{x}$ .

Now, the cost of diesel is Rs 35 per litre.

We need to ask ourselves, how many litres do we need for this journey? Well, if for every km traveled, $\dfrac{1}{400} \left(\dfrac{1000}{X} + X \right)$ litres were required, then how many litres will be required for the distance that we travel, which is $1600$ km. That is simply

$\Rightarrow 1600 \times \left( \dfrac{1}{400} \left(\dfrac{1000}{X} + X \right) \right)$

Thus, we need $1600 \times \left( \dfrac{1}{400} \left(\dfrac{1000}{X} + X \right) \right)$ litres. Now, the cost of traveling per litre is Rs 35. Therefore, the cost of traveling the distance that we traveled is

$\Rightarrow \left( \dfrac{1}{400} \left(\dfrac{1000}{X} + X \right) \right) \times 35 \times 1600$ .

The total cost as a function of $X$ is

$\Rightarrow f(X) =\left( \left( \dfrac{1}{400} \left(\dfrac{1000}{X} + X \right) \right) \times 35 \times 1600 \right) + \dfrac{1600 \times 125}{X}$

We want to minimize $f(X)$ . Taking the derivative, finding $f'(X)$ we can a local minima of $X = 10\sqrt{\dfrac{170}{7}} \approx 49.28$ . Therefore, our answer is $49.28 \approx \boxed{49}$ .

$Given\quad that\quad the\quad diesel\quad consumption\quad is\quad at\quad the\quad rate=\quad \frac { 1 }{ 400 } \left[ \frac { 100 }{ x } +x \right] \quad litres\quad per\quad km\\ Cost\quad of\quad diesel=\quad Rs\quad 35\quad per\quad litre\\ Payment\quad to\quad the\quad driver=\quad Rs\quad 125\quad per\quad hour.\\ Also\quad given\quad that\quad the\quad SUV\quad is\quad driven\quad at\quad the\quad speed\quad of\quad x\quad km\quad per\quad hour.\\ Total\quad cost(y)=\quad Cost\quad of\quad diesel\quad +\quad Payment\quad to\quad driver\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 1 }{ 400 } \left[ \frac { 100 }{ x } +x \right] \times 800\times 35\quad +\quad 125\times \left[ \frac { 800 }{ x } \right] \\ Now\quad differentiating\quad both\quad sides\quad in\quad the\quad above\quad equation\quad with\quad respect\quad to\quad x\quad and\quad equating\quad to\quad zero.\\ \frac { dy }{ dx } =\frac { -170000 }{ { x }^{ 2 } } \quad +\quad 70=0\quad \\ ⇒\quad \boxed { x=\quad 49\quad km\quad per\quad hour } .$

Minimize (800 35/400(1000/x+x))+125 (800/x) x~49.26

Total cost C = (125 800/X) + 800 35 [(100/x) + x]/400; dC/dX = -125 800/X^2 - 70*1000/X^2 + 70 =0; X ~ 49