equilateral triangle is increasing at a rate of √3 cm/min. find the rate at which its area increasing when its edge is 12 cm long

a = √3/4 s^2

da/dt = √3/2 s ds/dt

Now just plug in your numbers

3 cm which its area increasing when its edge is 12 cm long

huh? what does that mean??? Your alleged question is so poorly written it's incoherent. As I said, just plug in your numbers:

da/dt = √3/2 * 12 * √3 = 18

rude.

To find the rate at which the area of an equilateral triangle is increasing, we need to use the formulas for the area and the rate of change of the side length.

The formula for the area of an equilateral triangle is given by:
A = (sqrt(3)/4)*s^2
where A is the area and s is the side length.

Since we are given that the side length is increasing at a rate of √3 cm/min, we can differentiate both sides of the area formula with respect to time (t) to find the rate at which the area is increasing.

dA/dt = [(sqrt(3)/4)*2s] * ds/dt

Let's substitute the given values:
dA/dt = [(sqrt(3)/4)*2(12)] * √3

Simplifying this expression:
dA/dt = (3/2) * 24 * √3
dA/dt = 36 * √3 cm^2/min

Therefore, the rate at which the area of the equilateral triangle is increasing when its edge is 12 cm long is 36√3 cm^2/min.