calculus
posted by ssss .
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?

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. 
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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