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To solve this problem, let's go through the process step-by-step.

### Given Information
- Length \( \ell \) of the rectangle is increasing at a rate of \( \frac{d\ell}{dt} = 7 \, \text{cm/s} \).
- Width \( w \) of the rectangle is decreasing at a rate of \( \frac{dw}{dt} = -7 \, \text{cm/s} \).
- At a specific moment, \( \ell = 12 \, \text{cm} \) and \( w = 5 \, \text{cm} \).

We are asked to find:
1. The rate of change of the **area**.
2. The rate of change of the **perimeter**.
3. The rate of change of the **length of the diagonals**.

We’ll also determine if each of these quantities is increasing, decreasing, or constant.

---

### 1. Rate of Change of the Area

The area \( A \) of a rectangle is given by:
\[
A = \ell \times w
\]

To find \( \frac{dA}{dt} \), we use the product rule:
\[
\frac{dA}{dt} = \ell \frac{dw}{dt} + w \frac{d\ell}{dt}
\]

Substitute the given values:
\[
\frac{dA}{dt} = (12)(-7) + (5)(7)
\]
\[
\frac{dA}{dt} = -84 + 35 = -49 \, \text{cm}^2/\text{s}
\]

Since \( \frac{dA}{dt} = -49 \, \text{cm}^2/\text{s} \), the area is **decreasing**.

---

### 2. Rate of Change of the Perimeter

The perimeter \( P \) of a rectangle is given by:
\[
P = 2\ell + 2w
\]

Differentiate with respect to time \( t \):
\[
\frac{dP}{dt} = 2\frac{d\ell}{dt} + 2\frac{dw}{dt}
\]

Substitute the values:
\[
\frac{dP}{dt} = 2(7) + 2(-7)
\]
\[
\frac{dP}{dt} = 14 - 14 = 0 \, \text{cm/s}
\]

Since \( \frac{dP}{dt} = 0 \), the perimeter is **constant**.

---

### 3. Rate of Change of the Length of the Diagonals

The length \( D \) of the diagonal of a rectangle can be found using the Pythagorean theorem:
\[
D = \sqrt{\ell^2 + w^2}
\]

Differentiate \( D \) with respect to time:
\[
\frac{dD}{dt} = \frac{1}{2\sqrt{\ell^2 + w^2}} \cdot 2\left(\ell \frac{d\ell}{dt} + w \frac{dw}{dt}\right)
\]
\[
\frac{dD}{dt} = \frac{\ell \frac{d\ell}{dt} + w \frac{dw}{dt}}{\sqrt{\ell^2 + w^2}}
\]

Substitute the given values:
\[
\frac{dD}{dt} = \frac{(12)(7) + (5)(-7)}{\sqrt{12^2 + 5^2}}
\]
\[
= \frac{84 - 35}{\sqrt{144 + 25}}
\]
\[
= \frac{49}{\sqrt{169}} = \frac{49}{13} \, \text{cm/s}
\]

Since \( \frac{dD}{dt} = \frac{49}{13} \, \text{cm/s} \), the length of the diagonal is **increasing**.

---

### Summary

- **Area**: Decreasing at \( -49 \, \text{cm}^2/\text{s} \).
- **Perimeter**: Constant, rate of change is \( 0 \, \text{cm/s} \).
- **Diagonal**: Increasing at \( \frac{49}{13} \, \text{cm/s} \).

Would you like further details or have any questions?

---

#### Related Questions
1. How would the solution change if the width were increasing instead of decreasing?
2. What would happen to the area if the length and width increased at different rates?
3. How would we calculate the rate of change of the diagonal if the width were constant?
4. Can you explain why the perimeter remains constant in this case?
5. How is the rate of change of the diagonal derived from the Pythagorean theorem?

---

#### Tip:
For problems involving rates of change, remember to apply the chain rule carefully, especially when dealing with products or compositions of functions.

The length l of a rectangle is increasing at a rate of 7 cm/sec while the width w is decreasing at a rate of 7 cm/sec. When l = 12 cm and w = 5 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.

Explore a related rates problem where the length of a rectangle increases at 7 cm/sec while the width decreases at the same rate. Learn to find the rates of change for the area, perimeter, and diagonal, and determine if these quantities are increasing, decreasing, or constant. Suitable for Grades 10-12 students.

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