O lvl A Maths: Applications of Differentiation (Maxima and Minima)

Contributed by Javid
Adapted from an outdated textbook

A man rows 30 metres out to sea from point P on a straight coast. He reaches a point M such that MP is perpendicular to the coast. He then wishes to get as quickly as possible to a point Q on the coast 400 metres from P. If he can row at 40 m/min and cycle at 50 m/min, how far from P should he land?

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Answer:

Let the man land at the point R, x metres from P.

Using Pythagorean Theorem,
the distance he needs to row = $\sqrt{30^{2}+x^{2}} m$

The distance he needs to cycle = (400 – x) m

Total time taken:
$T = \frac{\sqrt{30^{2}+x^{2}}}{40}+\frac{400 - x}{50}$
$T=\frac{1}{40}\left(30^{2}+x^{2} \right)^{\frac{1}{2}}+8-\frac{x}{50}$

$\frac{dT}{dx}=\left(\frac{1}{2} \right)\left(\frac{1}{40} \right)\left(30^{2}+x^{2} \right)^{\frac{1}{2}}\left(2x \right)-\frac{1}{50}$
$\frac{dT}{dx}=\frac{x}{40\sqrt{30^2+x^{2}}}-\frac{1}{50}$

For a stationary value of T,
$\frac{dT}{dx}=0$
$\frac{x}{40\sqrt{30^2+x^{2}}}=\frac{1}{50}$
$50x=40\sqrt{30^2+x^2}$
$5x=4\sqrt{30^2+x^2}$
25x² = 16 (900 + x²)
9x² = 14400
x² = 1600
x = 40 (since x > 0)

When x = 39,
$\frac{dT}{dx}=\frac{39}{40\sqrt{30^2+39^2}}-\frac{1}{50}$ is negative.

When x = 41,
$\frac{dT}{dx}=\frac{41}{40\sqrt{30^2+41^2}}-\frac{1}{50}$ is positive.

=> the value of T = 40, is indeed a minimum point.

Thus, the man should land 40 m from P in order to reach Q as quickly as possible.

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O lvl A Maths: Applications of Differentiation (Maxima and Minima)