Question

A calculus optimization problem

Answer 1

Sure! Let's go through a calculus optimization problem step by step. Here’s a classic example:

**Problem Statement:**
You want to construct a rectangular garden that will be enclosed with a fence. You have 100 meters of fencing material available. What dimensions will maximize the area of the garden?

### Step 1: Define the Variables
Let:
- \( x \) = length of the garden in meters
- \( y \) = width of the garden in meters

### Step 2: Write the Constraints
The perimeter of the rectangular garden can be expressed as:
\[
P = 2x + 2y
\]
Given that you have 100 meters of fencing:
\[
2x + 2y = 100
\]
This simplifies to:
\[
x + y = 50
\]
or
\[
y = 50 - x
\]

### Step 3: Write the Area Function
The area \( A \) of the rectangle is:
\[
A = x \cdot y
\]
Substituting for \( y \):
\[
A = x(50 - x) = 50x - x^2
\]

### Step 4: Find the Critical Points
To find the maximum area, we need to take the derivative of the area function and set it to zero:
\[
\frac{dA}{dx} = 50 - 2x
\]
Setting the derivative equal to zero:
\[
50 - 2x = 0
\]
\[
2x = 50 \quad \Rightarrow \quad x = 25
\]

### Step 5: Find the Corresponding Width
Using the constraint \( y = 50 - x \):
\[
y = 50 - 25 = 25
\]

### Step 6: Verify that it’s a Maximum
To confirm that this critical point gives a maximum area, check the second derivative:
\[
\frac{d^2A}{dx^2} = -2
\]
Since the second derivative is negative, the function is concave down, indicating that we have a maximum.

### Conclusion
The dimensions that will maximize the area of the garden are \( x = 25 \) meters (length) and \( y = 25 \) meters (width), resulting in a square shape. The maximum area of the garden is:
\[
A = 25 \cdot 25 = 625 \text{ square meters.}
\]

Feel free to modify this problem or ask about a different type of optimization scenario!