SOLUTION: How to find point C on x-axis to get AC+BC is minimum with A(1,6) and B(6,1)? Thanks.

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Question 703463: How to find point C on x-axis to get AC+BC is minimum with A(1,6) and B(6,1)? Thanks.
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Let the desired point on the x axis be C(x,0)



Using the distance formula:

D = AC+BC

D = sqrt%28%28x-1%29%5E2%2B%286-0%29%5E2%29%2Bsqrt%28%28x-6%29%5E2%2B%281-0%29%5E2%29

D = sqrt%28%28x-1%29%5E2%2B36%29%2Bsqrt%28%28x-6%29%5E2%2B1%29

We need to find dD%2Fdx

Both those terms on the right are of the form 

u = sqrt%28v%5E2%2Bk%29

Differentiate that and you get

du%2Fdx = expr%28v%2Fsqrt%28v%5E2%2Bk%29%29%2Aexpr%28dv%2Fdx%29

So, using that on both terms: 

dD%2Fdx = %28x-1%29%2Fsqrt%28%28x-1%29%5E2%2B36%29 + %28x-6%29%2Fsqrt%28%28x-6%29%5E2%2B1%29

dD%2Fdx = %28x-1%29%2Fsqrt%28x%5E2-2x%2B1%2B36%29 + %28x-6%29%2Fsqrt%28x%5E2-12x%2B36%2B1%29

dD%2Fdx = %28x-1%29%2Fsqrt%28x%5E2-2x%2B37%29 + %28x-6%29%2Fsqrt%28x%5E2-12x%2B37%29

We set that = 0 to find minimum value:

%28x-1%29%2Fsqrt%28x%5E2-2x%2B37%29 + %28x-6%29%2Fsqrt%28x%5E2-12x%2B37%29 = 0

Clearing of fractions:

%28x-1%29sqrt%28x%5E2-12x%2B37%29 + %28x-6%29sqrt%28x%5E2-2x%2B37%29 = 0

%28x-1%29sqrt%28x%5E2-12x%2B37%29 = -%28x-6%29sqrt%28x%5E2-2x%2B37%29

Square both sides

(x-1)²(x²-12x+37) = (x-6)²(x²-2x+37)

(x²-2x+1)(x²-12x+37) = (x²-12x+36)(x²-2x+37)

Multiply that out and combine terms and get:

x4-14x³+62x²-86x+37 =  x4-14x³+97x²-516x+1332

That simplifies to:

35x²-430x+1295 = 0

Divide thru by 5: 

7x²-86x+259 = 0

That factors as

(7x-37)(x-7)=0

That has solutions x=37%2F7 and x=7.

We can show that x=7 is extraneous, since both terms of
dD%2Fdx are positive wnen x=7, so it isn't 0 there.

Thus x=37%2F7 is the only solution to dD%2Fdx = 0 

x=37%2F7 is between 1 and 6. It's easy to show that 
dD%2Fdx is negative when x=1 and positive when x=6, 
thus the desired point is C(37%2F7,0).

The graph showing the minimum of AC+BC drawn to scale is:



Edwin