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.