Problem 9 - Essential Calculus

Question

Find the length and width of a rectangle that has the given perimeter and a maximum area. Perimeter: 100 meters

Step-by-Step Solution

Answer

The dimensions that give the maximum area for a rectangle with a perimeter of 100 meters are a length of 25 meters and a width of 25 meters. This makes the rectangle a square, which is expected since among all rectangles with a given perimeter, a square has the maximum area.

Step 1: Write the formulas

Write down the formulas for the perimeter and area of a rectangle. The perimeter \(P\) is \(2l + 2w\) and the area \(A\) is \(lw\), where \(l\) and \(w\) are the rectangle's length and width respectively.

Step 2: Express the width in terms of the perimeter and length

We are given that the perimeter \(P\) is 100 meters, so we have \(2l + 2w = 100\). We can solve this equation for \(w\) to express it as a function of \(l\): \(w = (100 - 2l)/2 = 50 - l\).

Step 3: Express the area in terms of the length

Substitute \(w\) into the area formula to write \(A\) as a function of \(l\): \(A = l(50 - l)\).

Step 4: Find the maximum area

To find the maximum area, we need to maximize the function \(A(l) = l(50 - l)\). This is a parabola that opens downwards, and its maximum value occurs at its vertex. The x-coordinate of the vertex of a parabola given by \(y = a(x - h)^2 + k\) is \(h\), so in this case, \(l = 50/2 = 25\).

Step 5: Calculate width using the optimal length

Substitute \(l = 25\) into the equation for \(w\) that we found in Step 2 to find the width that gives the maximum area: \(w = 50 - 25 = 25\).

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is one of its most fundamental properties. It refers to the total distance around the outer edge of the rectangle. For a rectangle, the perimeter is calculated using the formula \( P = 2l + 2w \), where \( l \) is the length and \( w \) is the width. This formula helps in understanding how the length and width contribute equally to the overall perimeter of the shape.

• To find the perimeter, you simply add together the lengths of all four sides.
• If you know one dimension and the perimeter, you can easily find the other dimension using simple algebra.

In the context of our exercise, knowing that the perimeter is 100 meters allows us to set up an equation that expresses one dimension in terms of the other. This is helpful in optimization problems where constraints are given, such as maximizing the area.

Area of a Rectangle

The area of a rectangle is a measure of how much surface it covers. It is calculated by multiplying its length \( l \) by its width \( w \), expressed as \( A = l \cdot w \). Understanding the relationship between the area and its dimensions is crucial when tackling optimization problems.

• The formula \( A = l \cdot w \) indicates that if the length or width increases, so does the area, provided the other dimension stays constant.
• Maximizing the area given a fixed perimeter involves adjusting the dimensions appropriately.

In our exercise, to find the maximum area of the rectangle with a fixed perimeter, we manipulate the area formula by expressing width in terms of length, after understanding that \( w = 50 - l \). This transforms the problem into one of maximizing a quadratic function, a common approach in optimization.

Vertex of a Parabola

A parabola is a symmetric curve that can open upwards or downwards. The highest or lowest point on a parabola, depending on its orientation, is called the vertex. In problems of optimization, particularly those involving quadratic expressions like the area function \( A(l) = l(50 - l) \), recognizing the significance of the vertex helps in determining maximum or minimum values.

• A downward opening parabola reaches its maximum value at the vertex.
• The calculated vertex point gives the dimension required to maximize or minimize the value in question.

For our exercise, the expression for area \( A(l) = l(50 - l) \) forms a downward-opening parabola. To locate the vertex, we utilize the formula for the x-coordinate of a parabola's vertex \( x = -\frac{b}{2a} \). Plugging in values, we find \( l = 25 \), giving us not only the maximum area but also the idea of symmetry, implying \( w = 25 \) as well.