Calculus optimization help
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Calculus optimization help

[From: ] [author: ] [Date: 11-10-24] [Hit: ]
Now, all we have to do is find the angles:tan(t[1]) = 2/(4/3)tan(t[2]) = 1/(4 - (4/3))Does tan(t[1]) = tan(t[2])?tan(t[1]) = 6/4 = 3/2tan(t[2]) = 1 / (12/3 - 4/3) = 1 / (8/3) = 3/8Obviously, theres a problem.I cant see where I messed up, but I did notice that if I switched some values,......
x = (32 +/- sqrt(32 * 8)) / 6
x = (32 +/- sqrt(256)) / 6
x = (32 +/- 16) / 6
x = 48/6 , 16/6
x = 8 , 4/3

Now, x = 8 can't be a correct answer, since the overall length of x and 4 - x is 4 units

x = 4/3

There's the hard part; finding x. Now, all we have to do is find the angles:

tan(t[1]) = 2/(4/3)
tan(t[2]) = 1/(4 - (4/3))

Does tan(t[1]) = tan(t[2])?

tan(t[1]) = 6/4 = 3/2
tan(t[2]) = 1 / (12/3 - 4/3) = 1 / (8/3) = 3/8

Obviously, there's a problem. I can't see where I messed up, but I did notice that if I switched some values, the tangents are equal:

tan(t[1]) = 2 / (4 - x) = 2 / (4 - 4/3) = 2 / (8/3) = 2 * (3/8) = 3/4
tan(t[1]) = 1 / (4/3) = 1 * (3/4) = 3/4

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And that's where I went wrong. 16/6 obviously doesn't reduce to 4/3.

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why use dratted calculus where a little creativity will do ?


|A
|
|
|............................... ..................|B
|............................... ..................|
|................................ .................|
X------------------------------P --------------- |Y
................................. ................. |
................................. ................. |
................................. ................. |B'

draw a point B' drawn for an equal distance behind the mirror
for shortest distance, APB' must be a straight line
AXP & B'YP (& hence BYP) are similar right angle triangles,
and you should now be able to get the asked for result easily
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keywords: help,optimization,Calculus,Calculus optimization help
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