Each side of a square is increasing at a rate of 3 cm/s. At what rate is the area of the square increasing when the area of the square is 25 cm^2?

The rate at which the area of a square is increasing can be found by taking the derivative of the function that represents the area of the square with respect to time.

If the side length of the square is represented by the variable $s$, the area of the square can be represented by the function $A(s) = s^2$. The rate at which the area is changing with respect to time can be found by taking the derivative of this function with respect to time, which gives us the following:

$\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$

Substituting the given value of 3 cm/s for the rate at which the side length is increasing, we get the following:

$\frac{dA}{dt} = 2s \cdot 3$ cm/s

Substituting the given value of 25 cm^2 for the area of the square, we can find the value of s by solving the equation $A(s) = s^2 = 25$ cm^2, which gives us $s = 5$ cm. Substituting this value for s in the equation for the rate at which the area is changing, we get the following:

$\frac{dA}{dt} = 2 \cdot 5$ cm $\cdot 3$ cm/s = 30 cm^2/s

Therefore, when the area of the square is 25 cm^2, the rate at which the area is increasing is 30 cm^2/s.