Finding the minimum length of a median in a triangle

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Oh, you're correct! I forgot the actual question lol. :) Yes, I was solving for x, not for AD. Many thanks mege and ILS! You two are brilliant!
 
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Edit Realised you can't solve it this way, sorry will come back.

Yes, reading your post I was like that...

But thanks for your feedback. As I've said before, you and ILS in the same thread is a total... ;)

So here is the amended version using implicit differentiation.

Let y = AD2

[tex]\begin{array}{l}<br /> y = {x^2} + {\left( {4 - 2x} \right)^2} \\ <br /> y = {x^2} + 4{x^2} - 16x + 16 \\ <br /> y = 5{x^2} - 16x + 16 \\ <br /> dy = \left( {10x - 16} \right)dx \\ <br /> \frac{{dy}}{{dx}} = \left( {10x - 16} \right) = 0 \\ <br /> x = 1.6 \\ <br /> \end{array}[/tex]

Implicit differentiation is a good trick to learn as it can significantly reduce the work.

go well