Posted by ssss on Thursday, October 18, 2007 at 9:18pm.
The diagonal of a square is increasing at a rate of 3 inches per minute. When the area of the square is 18 square inches, how fast(in inches per minute) is the perimeter increasing?

calculus  bobpursley, Thursday, October 18, 2007 at 9:44pm
The 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 don't see it vary with the size of the square. check my thinking.

calculus  Reiny, Thursday, October 18, 2007 at 9:56pm
bobpursely 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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