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Let $x$ be the side of the equilateral triangle and $A$ be its area. We are given that $\frac{dx}{dt} = 3$ cm/s. We want to find $\frac{dA}{dt}$ when $x = 15$ cm.
The area of an equilateral triangle with side $x$ is given by: $$A = \frac{\sqrt{3}}{4}x^2$$
Differentiating both sides of the equation with respect to $t$, we get: $$\frac{dA}{dt} = \frac{d}{dt} \left( \frac{\sqrt{3}}{4}x^2 \right)$$ $$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2x \cdot \frac{dx}{dt}$$ $$\frac{dA}{dt} = \frac{\sqrt{3}}{2}x \frac{dx}{dt}$$
We are given that $x = 15$ cm and $\frac{dx}{dt} = 3$ cm/s. Substituting these values into the equation, we get: $$\frac{dA}{dt} = \frac{\sqrt{3}}{2} (15) (3)$$ $$\frac{dA}{dt} = \frac{45\sqrt{3}}{2}$$
The rate at which the area is increasing is $\frac{45\sqrt{3}}{2}$ cm$^2$/s.
Final Answer: $\frac{45\sqrt{3}}{2}$ cm$^2$/s
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