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September 16, 2014

September 16, 2014

Posted by **Furzeen** on Tuesday, November 29, 2011 at 5:58am.

- Math(module 4 Calculus) -
**Reiny**, Tuesday, November 29, 2011 at 7:47amLet the distance from A to his landing place be x

then from there it would be another 5-x miles to his house.

let the distance he rows be d

d^2 = x^2 + 2^2

d = (x^2 + 4)^(1/2)

time for the rowing part = (x^2 + 4)^(1/2) /3

time for the walking part = (5-x)/5

Time = (x^2 + 4)^(1/2) /3 + (5-x)/5

d(Time)/dx = (1/6)(x^2 + 4)^(-1/2) (2x) - 1/5

x/(3√(x^2+4) ) - 1/5

= 0 for a minimum of Time.

x/(3√(x^2+4) ) - 1/5

5x = 3√(x^2 + 4)

25x^2 = 9(x^2+4)

25x^2 = 9x^2 + 36

16x^2 = 36

4x = 6

x = 6/4

sub that into Time = .. and you got it

( I got 1 hour and 32 minutes)

check my arithmetic

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