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February 1, 2015

February 1, 2015

Posted by **Chris** on Friday, December 16, 2011 at 4:08am.

- Calculus -
**Damon**, Friday, December 16, 2011 at 5:08amx y = 1

c = 3 x + x + 2 y = 4 x + 2 y

c = 4 x + 2/x

dc/dx = 4 - 2/x^2

= 0 for minimum

4 = 2/x^2

x^2 = 1/2

x = 1/sqrt 2 = sqrt 2/2 along road

y = sqrt 2

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