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Let's solve each question step by step.

### Question 1
> **Find two positive numbers whose product is 36 and whose sum is a minimum.**

Let the two positive numbers be \( x \) and \( y \). We know:
1. \( x \cdot y = 36 \)
2. We need to minimize the sum, \( S = x + y \).

To express \( y \) in terms of \( x \), use the first equation:
\[
y = \frac{36}{x}
\]

Now, substitute into \( S \):
\[
S = x + \frac{36}{x}
\]

To find the minimum sum, take the derivative of \( S \) with respect to \( x \) and set it to zero:
\[
\frac{dS}{dx} = 1 - \frac{36}{x^2} = 0
\]

Solving this:
\[
1 = \frac{36}{x^2} \Rightarrow x^2 = 36 \Rightarrow x = 6
\]

Since \( y = \frac{36}{x} \), we get:
\[
y = \frac{36}{6} = 6
\]

Thus, the two numbers are **6 and 6**.

---

### Question 2
> **Find the dimensions of a rectangle (in m) with a perimeter of 84 m whose area is as large as possible.**

Let the length be \( l \) and the width be \( w \). We know:
1. The perimeter is \( 2l + 2w = 84 \), so \( l + w = 42 \).
2. We need to maximize the area, \( A = l \cdot w \).

Express \( w \) in terms of \( l \) using the perimeter:
\[
w = 42 - l
\]

Now, substitute into \( A \):
\[
A = l \cdot (42 - l) = 42l - l^2
\]

To find the maximum area, take the derivative of \( A \) with respect to \( l \) and set it to zero:
\[
\frac{dA}{dl} = 42 - 2l = 0
\]

Solving this:
\[
2l = 42 \Rightarrow l = 21
\]

Since \( w = 42 - l \), we get:
\[
w = 42 - 21 = 21
\]

Therefore, the dimensions of the rectangle are **21 m by 21 m**.

Would you like further details on any part of the solution?

---

### Follow-up Questions
1. Why does minimizing the sum in Question 1 lead to two equal numbers?
2. How does setting the derivative to zero help find maximum or minimum values?
3. Could we solve these problems using a different method, like graphing?
4. Why does a square (21 m by 21 m) maximize the area for a given perimeter?
5. How would the solution change if we were asked for integer-only dimensions?

### Tip
When optimizing, turning constraints into equations is the first key step. This helps simplify the problem into a single-variable function for easier calculation.

Learn to solve optimization problems, including finding two numbers with a minimum sum for a given product and maximizing the area of a rectangle with a fixed perimeter. Ideal for high school students studying calculus and algebra.

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