# Problem 506 | Friction

Infinitesimal Part (c)

Here is one alternative solution in minimizing $P=\dfrac{160}{\cos \alpha + 0.4 \sin \alpha}$.

To minimize $P$, the denominator should be maximize. So we consider $\cos \alpha + 0.4 \sin \alpha$.

1) Trigonometric identity $a\sin x + b\cos x = \sqrt{a^2 + b^2} \sin \left(\tan^{-1} \dfrac{b}{a} + x \right)$

$\cos \alpha + 0.4\sin \alpha = \sqrt{1 + 0.4^2} \sin \left(\tan^{-1} \dfrac{1}{0.4} + \alpha \right)$

But since $-1 \leq \sin x \leq 1$, $\sin \left(\tan^{-1} \dfrac{1}{0.4} + \alpha \right)$ has $1$ as the maximum value. Thus,
$\cos \alpha + 0.4 \sin \alpha = \sqrt{1 + 0.4^2}$

Hence, the minimum value of $P$ is $P=\dfrac{160}{\sqrt{1.16}}=\boxed{148.56 \, \mathrm{lb}}$

2) Cauchy-Schwarz Inequality, $(a^2 + b^2)(c^2+d^2) \geq (ac + bd)^2$

We have,
\begin{eqnarray*}
(1^2 + 0.4^2)(\cos^2 \alpha + \sin^2 \alpha) &\geq& (\cos \alpha + 0.4\sin \alpha)^2\\
(\cos \alpha + 0.4\sin \alpha)^2 &\leq& (1.16)(1)\\
-\sqrt{1.16} \leq &(\cos \alpha + 0.4 \sin \alpha)& \leq \sqrt{1.16}
\end{eqnarray*}
Which means the maximum value of $\cos \alpha + 0.4 \sin \alpha = \sqrt{1.16}$ and it goes back to the fact that the minimum value of $P$ is $\boxed{148.56 \, \mathrm{lb}}$

Jhun Vert Thank you for sharing sir. The fastest way to solve this problem is graphical, a practical way for time constraint exams. One may consider this figure to find the Pmin. CE student You can use Lami's Theorem also

Jhun Vert Yes, that is a straightforward solution. Thank you.

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