To solve this problem step by step:

We are given:

1. Area of the rectangle is constant: $A = 420 \, \text{square feet}$.
2. Length increases at a rate: $\frac{dL}{dt} = 3 \, \text{ft/min}$.
3. We need to find the rate of change of the perimeter, $\frac{dP}{dt}$, when the length $L = 35 \, \text{ft}$.

Step 1: Relations among variables

Let:

• $L$ = length,
• $W$ = width.

The area of the rectangle is: $A = L \cdot W.$ Since $A = 420$ is constant, we get: $L \cdot W = 420.$

The perimeter of the rectangle is: $P = 2L + 2W.$

We need $\frac{dP}{dt}$, the rate of change of $P$.

Step 2: Differentiate the area equation with respect to time $t$

Start with: $L \cdot W = 420.$ Differentiate both sides with respect to $t$ (using the product rule): [ \frac{d}{dt}[L \cdot W] = \frac{d}{dt}[420]. ] This gives: $\frac{dL}{dt} \cdot W + L \cdot \frac{dW}{dt} = 0.$

Rearranging for $\frac{dW}{dt}$, the rate of change of $W$: $\frac{dW}{dt} = -\frac{\frac{dL}{dt} \cdot W}{L}.$

Step 3: Find $W$ when $L = 35$

From the area equation $L \cdot W = 420$, substitute $L = 35$: $35 \cdot W = 420.$ Solve for $W$: $W = \frac{420}{35} = 12 \, \text{ft}.$

Step 4: Find $\frac{dW}{dt}$

We now substitute into the differentiated area equation:

• $\frac{dL}{dt} = 3 \, \text{ft/min}$,
• $L = 35$,
• $W = 12$.

From: $\frac{dW}{dt} = -\frac{\frac{dL}{dt} \cdot W}{L},$ we get: $\frac{dW}{dt} = -\frac{3 \cdot 12}{35}.$ Simplify: $\frac{dW}{dt} = -\frac{36}{35} \, \text{ft/min}.$

Step 5: Differentiate the perimeter equation

The perimeter is: $P = 2L + 2W.$ Differentiate both sides with respect to $t$: $\frac{dP}{dt} = 2 \frac{dL}{dt} + 2 \frac{dW}{dt}.$

Substitute the known values:

• $\frac{dL}{dt} = 3$,
• $\frac{dW}{dt} = -\frac{36}{35}$.

$\frac{dP}{dt} = 2(3) + 2\left(-\frac{36}{35}\right).$ Simplify: $\frac{dP}{dt} = 6 - \frac{72}{35}.$

Convert $6$ to have a denominator of 35: $6 = \frac{210}{35}.$ Thus: $\frac{dP}{dt} = \frac{210}{35} - \frac{72}{35}.$ Combine the fractions: $\frac{dP}{dt} = \frac{138}{35}.$

Step 6: Final Answer

The perimeter is increasing at a rate of: $\frac{138}{35} \approx 3.94 \, \text{ft/min}.$

Clarifications or Further Questions?

Let me know if you need more details! Here are related questions to explore further:

1. How does the rate of change of the area relate to length and width?
2. What happens if the area is not constant but increasing at a certain rate?
3. How do you find the rate of change of the diagonal of the rectangle?
4. Can this method be extended to solve problems involving other shapes, like a circle?
5. How does the rate of change of the width compare to the rate of change of the length?

Tip: When dealing with related rates, always differentiate the given equation carefully, applying the chain rule where needed!