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November 27, 2014

November 27, 2014

Posted by **ssss** on Thursday, October 18, 2007 at 9:18pm.

- calculus -
**bobpursley**, Thursday, October 18, 2007 at 9:44pmThe sides are .707 of the diagonal, and area is s squared, so

Area= .5 diagonal^2

or diagonal= sqrt (2*area)

Perimeter= 4s=4(.707)diagonal

dP/dt= 4*.707 * diagonal/dt

= 4*.707 3in/min

I dont see it vary with the size of the square. check my thinking.

- calculus -
**Reiny**, Thursday, October 18, 2007 at 9:56pmbobpursely is right, the area of 18 in^2 has nothing to do with it.

Here is how I did it

D^2 = 2s^2

D = √2 s

dD/dt = √2 ds/dt

3 = √2 ds/dt ------> ds/dt = 3/√2

P = 4s

dP/dt = 4ds/dt=4(3/√2) = 12/√2 in/min

which is the same as bobpursley's answer

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