Maximizing a Sum of Distances

[turophile]

Here's the problem:

Locate the point P = (x, y) on the quarter-circle x[sup:3cab5ydf]2[/sup:3cab5ydf] + y[sup:3cab5ydf]2[/sup:3cab5ydf] = 1 (x ? 0, y ? 0) so that the sum of the distances from P to the y-axis and to the point (1, 0) shall be a maximum. What is the maximum?

Here's my (incorrect) solution:

The distance from P to the y-axis is x, and the distance from P to (1, 0) is sqrt[(1 - x)[sup:3cab5ydf]2[/sup:3cab5ydf] + y[sup:3cab5ydf]2[/sup:3cab5ydf]]. Let s be the sum of the distances. We know y = sqrt(1 - x[sup:3cab5ydf]2[/sup:3cab5ydf]). So s = x + sqrt[(1 - x)[sup:3cab5ydf]2[/sup:3cab5ydf] + [sqrt(1 - x[sup:3cab5ydf]2[/sup:3cab5ydf])][sup:3cab5ydf]2[/sup:3cab5ydf]] = x + sqrt(1 - 2x + x[sup:3cab5ydf]2[/sup:3cab5ydf] + 1 - x[sup:3cab5ydf]2[/sup:3cab5ydf]) = x + sqrt(2 - 2x). Then

s' = 1 + (1/2) ? 1/sqrt(2 - 2x) ? (- 2) = 1 - 1/sqrt(2 - 2x) ? 0

? 1/sqrt(2 - 2x) = 1
? sqrt(2 - 2x) = 1
? 2 - 2x = 1
? 2x = 2
? x = 1

My graph of s (and the textbook's answer) say s is a maximum at x = 1/2, so I've made a mistake somewhere, but I can't seem to find it.

 

[Ryan Rigdon]

when you did

? 1/sqrt(2 - 2x) = 1
? sqrt(2 - 2x) = 1
? 2 - 2x = 1
? 2x = 2
? x = 1

this is what you should have done

2-2x=1
-2x=-1

x = 1/2

hope this helps.

 

[tkhunny]

...and you were doing SO WELL.

Start here, again: 2 - 2x = 1

Then remember that 2 - 1 = 1.

 

[turophile]

We'll that was embarrassing! Thanks to both of you for bringing the error to my attention.