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(A) Given that the length $x$ is decreasing at the rate of $3 	ext{ cm/min}$,so $\frac{dx}{dt} = -3 	ext{ cm/min}$.Given that the width $y$ is incre

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$2 	ext{ cm}^2/	ext{min}$

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(A) Given that the length $x$ is decreasing at the rate of $3 	ext{ cm/min}$,so $\frac{dx}{dt} = -3 	ext{ cm/min}$.Given that the width $y$ is increasing at the rate of $2 	ext{ cm/min}$,so $\frac{dy}{dt} = 2 	ext{ cm/min}$.The area $A$ of the rectangle is given by $A = x \cdot y$.Differentiating both sides with respect to $t$,we get $\frac{dA}{dt} = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt}$.Substituting the given values $x = 10 	ext{ cm}$,$y = 6 	ext{ cm}$,$\frac{dx}{dt} = -3 	ext{ cm/min}$,and $\frac{dy}{dt} = 2 	ext{ cm/min}$:$\frac{dA}{dt} = (-3)(6) + (10)(2) = -18 + 20 = 2 	ext{ cm}^2/	ext{min}$.Thus,the rate of change of the area is $2 	ext{ cm}^2/	ext{min}$.

The length $x$ of a rectangle is decreasing at the rate of $3 \text{ cm/min}$ and the width $y$ is increasing at the rate of $2 \text{ cm/min}$. When $x = 10 \text{ cm}$ and $y = 6 \text{ cm}$,find the rate of change of the area of the rectangle.

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