A problem to maximize (optimize) the area of a rectangle with a constant perimeter is presented. An analytical method, based on the derivatives of a function and calculus theorems, is developed in order to find an analytical solution to the problem.
In what follows, \( C \) represents the constant of integration where applicable.
Problem
You decide to construct a rectangle of perimeter \( 400 \; \text{mm} \) and maximum area. Find the length and the width of the rectangle.
Show Solution to the Problem
Let \( x \) (width, distance \( DC \)) and \( y \) (length, distance \( DA \)) be the dimensions of the rectangle. The area \( A \) and perimeter \( P \) of the rectangle are given by:
\[ A = xy \] \[ P = 400 = 2x + 2y \]Solve the perimeter equation for \( y \):
\[ y = 200 - x \]Substitute \( y = 200 - x \) into the area formula to obtain area as a function of \( x \):
\[ A(x) = x(200 - x) = -x^2 + 200x \]The domain of function \( A(x) \) is the closed interval \( [0, 200] \) since \( x \ge 0 \) and \( y = 200 - x \ge 0 \) (which implies \( x \le 200 \)).
To find the value of \( x \) that maximizes \( A \), we calculate the first derivative \( \dfrac{dA}{dx} \):
\[ \frac{dA}{dx} = -2x + 200 \]Setting \( \dfrac{dA}{dx} = 0 \) to find critical points:
\[ -2x + 200 = 0 \implies x = 100 \]At the endpoints of the domain, \( A(0) = 0 \) and \( A(200) = 0 \). The second derivative is:
\[ \frac{d^2A}{dx^2} = -2 \]Since the second derivative is negative (\( -2 < 0 \)), by the second derivative test, the critical point \( x = 100 \) corresponds to a local maximum (which is an absolute maximum on the closed interval).
The corresponding length is \( y = 200 - 100 = 100 \; \text{mm} \), and the maximum area is \( A(100) = 100 \times 100 = 10000 \; \text{mm}^2 \). Thus, selecting a rectangle of width \( x = 100 \; \text{mm} \) and length \( y = 100 \; \text{mm} \) (a square) yields the maximum possible area.
Exercises
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Exercise 1: Solve the same problem as above but with a perimeter equal to \( 500 \; \text{mm} \).
Show Solution to Exercise 1
Using the perimeter formula \( 2x + 2y = 500 \implies y = 250 - x \). The area function is:
\[ A(x) = x(250 - x) = -x^2 + 250x \]Taking the first derivative:
\[ \frac{dA}{dx} = -2x + 250 = 0 \implies x = 125 \; \text{mm} \]The length is \( y = 250 - 125 = 125 \; \text{mm} \).
Conclusion: Width \( x = 125 \; \text{mm} \) and length \( y = 125 \; \text{mm} \) (a square).