\begin{tabular}{l|l} The length \( \ell \) of a rectangle is & The rate of change of the \\ decreasing at a rate of \( 3 \mathrm{~cm} / \mathrm{sec} \) & area of the rectangle is \\ while the width \( w \) is increasing at a & \( \mathrm{cm}^{2} / \mathrm{sec} \) \\ rate of \( 3 \mathrm{~cm} / \mathrm{sec} \). When \( \ell=12 \mathrm{~cm} \) & (Simplify your answer.) \\ and \( w=5 \mathrm{~cm} \), find the rates of & \\ change of the area, the perimeter, \\ and the lengths of the diagonals of \\ the rectangle. Determine which of & \\ these quantities are increasing, & \\ decreasing, or constant. \end{tabular}

To find the rates of change of the area, perimeter, and lengths of the diagonals of the rectangle, we can use related rates. **Area \( A \)** of a rectangle is given by \( A = \ell \times w \). The rate of change of area \( \frac{dA}{dt} \) can be found using the product rule: \[ \frac{dA}{dt} = \ell \cdot \frac{dw}{dt} + w \cdot \frac{d\ell}{dt} \] Substituting \(\ell = 12 \, \text{cm}\), \(w = 5 \, \text{cm}\), \(\frac{d\ell}{dt} = -3 \, \text{cm/s}\), and \(\frac{dw}{dt} = 3 \, \text{cm/s}\): \[ \frac{dA}{dt} = 12 \cdot 3 + 5 \cdot (-3) = 36 - 15 = 21 \, \text{cm}^2/\text{s} \] Thus, the area is increasing at a rate of \( 21 \, \text{cm}^2/\text{s} \). **Perimeter \( P \)** of the rectangle is given by \( P = 2(\ell + w) \). The rate of change of perimeter \( \frac{dP}{dt} \): \[ \frac{dP}{dt} = 2\left(\frac{d\ell}{dt} + \frac{dw}{dt}\right) \] Substituting the rates: \[ \frac{dP}{dt} = 2\left(-3 + 3\right) = 2 \cdot 0 = 0 \, \text{cm/s} \] Thus, the perimeter is constant. **Diagonal \( d \)** of the rectangle is given by \( d = \sqrt{\ell^2 + w^2} \). To find \( \frac{dd}{dt} \), we use: \[ \frac{dd}{dt} = \frac{1}{2\sqrt{\ell^2 + w^2}} \cdot (2\ell \cdot \frac{d\ell}{dt} + 2w \cdot \frac{dw}{dt}) \] This simplifies to: \[ \frac{dd}{dt} = \frac{\ell \cdot \frac{d\ell}{dt} + w \cdot \frac{dw}{dt}}{\sqrt{\ell^2 + w^2}} \] Calculating it: \[ \frac{dd}{dt} = \frac{12 \cdot (-3) + 5 \cdot 3}{\sqrt{12^2 + 5^2}} = \frac{-36 + 15}{\sqrt{144 + 25}} = \frac{-21}{\sqrt{169}} = -\frac{21}{13} \approx -1.62 \, \text{cm/s} \] Thus, the diagonal is decreasing at a rate of about \( 1.62 \, \text{cm/s} \). To summarize: - Area is increasing at \( 21 \, \text{cm}^2/\text{s} \) (increasing). - Perimeter is constant (no change). - Diagonal is decreasing at approximately \( 1.62 \, \text{cm/s} \) (decreasing).