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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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$-2 	ext{ cm/min}$

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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 perimeter $P$ of a rectangle is given by the formula $P = 2(x + y)$.To find the rate of change of the perimeter with respect to time $t$,we differentiate $P$ with respect to $t$:$\frac{dP}{dt} = 2 \left( \frac{dx}{dt} + \frac{dy}{dt} ight)$.Substituting the given values:$\frac{dP}{dt} = 2(-3 + 2) = 2(-1) = -2 	ext{ cm/min}$.Thus,the rate of change of the perimeter is $-2 	ext{ cm/min}$.

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 perimeter.

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