At 32 cm^2/s rate the area of the square is increasing when the area of the square is 64 [tex]cm^{2}[/tex]
As per given problem each side of a square is increasing at a rate of 2 cm/s. let side of square is x.so the area of it is A= [tex]x^{2}[/tex].now differentiating w.r.t time A
[tex]\frac{dA}{dt}[/tex]= 2x[tex]\frac{dx}{dt}[/tex]
when A=64 [tex]cm^{2}[/tex] x=8 cm
also given [tex]\frac{dx}{dt}[/tex]=2cm/s
so [tex]\frac{dA}{dt}[/tex]=2×2×8=32
So at 32 cm^2/s rate the area of the square is increasing when the area of the square is 64 [tex]cm^{2}[/tex]
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