Maximizing the Area of a Rectangle Given a Fixed Perimeter

Introduction to the Problem

Given a fixed perimeter of a rectangle, the question arises: How do we maximize the area? This problem is often encountered in various real-world applications such as designing gardens, layout planning, and even in engineering. Let's explore the solution step-by-step and understand the mathematical principles behind it.

Understanding the Perimeter and Area Formulas

The perimeter (P) of a rectangle is given by:

(P 2L 2W)

where (L) is the length and (W) is the width. For a given perimeter of 440 cm, the formula simplifies to:

(2L 2W 440)

Simplifying further, we get:

(L W 220)

Express the Area in Terms of One Variable

The area (A) of a rectangle is given by:

(A L times W)

From the perimeter equation, we can express (W) in terms of (L):

(W 220 - L)

Substituting this into the area formula, we get:

(A L times (220 - L) 220L - L^2)

Find the Maximum Area

The area function (A -L^2 220L) is a quadratic equation that opens downwards (since the coefficient of (L^2) is negative). The maximum area occurs at the vertex of the parabola. The vertex form of a quadratic equation (ax^2 bx c) can be found using the formula:

(L -frac{b}{2a})

Here, (a -1) and (b 220). Plugging these values in, we get:

(L -frac{220}{2 times -1} 110)

Calculate the Corresponding Width

Substitute (L 110) back to find (W):

(W 220 - L 220 - 110 110)

Therefore, the dimensions that maximize the area are:

Length (L 110) cm Width (W 110) cm

Thus, the rectangle is actually a square with side length 110 cm, and the maximum area is:

(110 times 110 12100) cm2

Real-World Application and Integer Constraints

For practical purposes, where the dimensions must be integers:

Half of the perimeter (P) is

(frac{P}{2} 220)

Therefore, the area (A) would measure between:

(1 times 219 219) cm2 (109 times 111 12099) cm2

If including decimal places, the largest area a rectangle can have with the perimeter of 440 cm is just short of 12100 cm2.

Conclusion

The optimal dimensions for maximizing the area of a rectangle given a fixed perimeter of 440 cm are 110 cm by 110 cm, resulting in a maximum area of 12100 cm2. This example demonstrates the power of quadratic optimization in practical scenarios.