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(D) Let $A$ be the area,$b$ be the breadth,and $\ell$ be the length of the rectangle.Given: $\frac{dA}{dt} = -5$,$\frac{d\ell}{dt} = 2$,and $\frac{db}

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$3$

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(D) Let $A$ be the area,$b$ be the breadth,and $\ell$ be the length of the rectangle.Given: $\frac{dA}{dt} = -5$,$\frac{d\ell}{dt} = 2$,and $\frac{db}{dt} = -3$.We know that the area of a rectangle is $A = \ell 	imes b$.Differentiating both sides with respect to time $t$,we get:$\frac{dA}{dt} = \ell \cdot \frac{db}{dt} + b \cdot \frac{d\ell}{dt}$.Substituting the given values:$-5 = \ell(-3) + b(2)$.$-5 = -3\ell + 2b$.When the breadth $b = 2 \, m$,we substitute this into the equation:$-5 = -3\ell + 2(2)$.$-5 = -3\ell + 4$.$3\ell = 4 + 5$.$3\ell = 9$.$\ell = 3 \, m$.Thus,the length of the rectangle is $3 \, m$.

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