Why Does My Snell's Law Demonstration Fail Using a Linear Function Approach?

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physics user1
I can't figure out why my demonstration of snell's law fails, that's the demonstration: (I used a photo)
I think it fails because the function t (HO) represents a line and so the concept of minimum is not defined, when I take the derivative and equal it to 0 I'm considering the case when the line is parallel to the x-axis (the first derivative gives me the angular coefficient and when I equal it to 0 the line is parallel) (considering as y the time and as x HO)

Is that the problem? ( I know the real demonstration but i want to understand why this variant doesn't work)
 
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  • Handwritten derivation on lined paper: a diagram with points A, O, B, H, K and angles theta1 and theta2 between two media n1 and n2, followed by equations for total time t(HO) and dt/d(HO) = 0 leading to n1 sin theta2 = n2 sin theta1, crossed out against n2 sin theta1 = n1 sin theta2.
    1451434372187-269526573.jpg
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theodoros.mihos said:
##AO\sin\theta_1=AH##

But... θ1 is the angle whit the vertical
 
You are right. May be better a simpler geometry.
Let ##A(x_1,y_1)## and ##B(x_2,y_2)## the end points and ##O(x,0)## the point than we need. Write the total time and take derivation ##d/dx(t)=0##.
 
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yep it is preferable to use simple geometry to demonstrate