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If the petrol burnt in driving a motor boat varies as the cube of the velocity, then the speed (in $\mathrm{km} /$ hour) of the boat going against a water flow of $C \mathrm{kms} /$ hour so that the quantity of petrol burnt is minimum is
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The correct answer is:
$\frac{3 C}{2}$
Let $V>C$ the speed of motor boat, relative to water. In fact, relative to land, the motor boat moves with the speed $V-C$.
The motor boat covers a distance $S$ in $\frac{S}{V-C} \mathrm{~h}$ and consumption is
$f(v)=k v^3 \cdot \frac{S}{V-C}$, where $k$ is a constant.
The minimum of consumption is attained for
$$
\begin{aligned}
& f^{\prime}(v)=0 \\
& \Rightarrow \quad k S \frac{(v-C) 3 v^2-v^3(1)}{(v-C)^2}=0 \\
& \Rightarrow \quad 3 v^3-3 v^2 C-v^3=0 \\
& \Rightarrow \quad 2 v^3-3 v^2 C=0 \Rightarrow v=\frac{3 C}{2} \text {. } \\
&
\end{aligned}
$$
The motor boat covers a distance $S$ in $\frac{S}{V-C} \mathrm{~h}$ and consumption is
$f(v)=k v^3 \cdot \frac{S}{V-C}$, where $k$ is a constant.
The minimum of consumption is attained for
$$
\begin{aligned}
& f^{\prime}(v)=0 \\
& \Rightarrow \quad k S \frac{(v-C) 3 v^2-v^3(1)}{(v-C)^2}=0 \\
& \Rightarrow \quad 3 v^3-3 v^2 C-v^3=0 \\
& \Rightarrow \quad 2 v^3-3 v^2 C=0 \Rightarrow v=\frac{3 C}{2} \text {. } \\
&
\end{aligned}
$$
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