Best Answer:
Using Ada's house as the point of reference:

Ian is coming towards her house at 8 kph from 20 km away,

so the distance to Ada's house is 20 - 8t.

Ada is heading off at a right angle at 6 kph,

so the distance away from her house is 6t.

Since they are perpendicular angles from the house,

we can use the pythagorean theorem to calculate the distance:

Distance, d, is square root of (20-8t)^2 + (6t)^2

= sqrt (100t^2 - 320t + 400)

Now, it's tempting to think that they are at the minimum distance

when they start (t=0): 20km

or when they end (t=2.5): 15km

or when they are equidistant from Ada's house:

20 - 8t = 6t -> t = 1.43 hrs -> 12.1 km

But let's see if we can't use some calculus.

The minimum or maximum of an equation can be found

by setting the derivative equal to zero

(since the slope is zero).

We can ignore the square root since the maximum with it

will be at the same value for t as without it.

(if you do it out, you'll see it's the same...

the derivative is:

0.5 (200t-320) / sqrt (100t^2 -320t +400))

So ... derivative of 100t^2 + 320t + 400 equals 0

200t - 320 = 0 -> t = 1.6 hours

Ian is then 7.2 km away and Ada is 9.6 km away

For a distance, d, of 12 km.

We know that at t = 1.43 hours, d = 12.1 km

(from above conjecture at a solution).

What about at 1.7 hours?

Ian would be 6.4 km away

Ada would be 10.2 km away

They would be 12.04 km apart.

What about at 1.5 hours?

Ian would be 9 km away

Ada would be 8 km away

They would be 12.04 km apart.

So, I think we've proven your answer

though without doing the full derivative

or a second derivative

(with which we could prove whether

it was a minimum or maximum).

Depending upon the level of your class,

you may be asked upon to solve with

the full derivative equal to zero (min/max)

and the second derivative's value positive (minimum).

Ian and Ada are closest after 1 hour 16 minutes (1.6 hours)

when they are exactly 12 km apart.

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