of A Rectangle with Given Perimeter

A problem to maximize (optimization) the area of a rectangle with a constant perimeter is presented. An interactive applet (you need Java in your computer) is used to understand the problem. Then an analytical method, based on the derivatives of a function and some calculus theorems, is developed in order to find an analytical solution to the problem.

## ProblemYou decide to construct a rectangle of perimeter 400 mm and maximum area. Find the length and the width of the rectangle.Solution to the Problem
We now look at a solution to this problem using derivatives and other calculus concepts. Let x ( = distance DC) be the width of the rectangle and y ( = distance DA)its length, then the area A of the rectangle may written: The perimeter may be written as Solve equation 400 = 2x + 2y for y We now now substitute y = 200 - x into the area A = x*y to obtain . Area A is a function of x. As you change the width x in the applet, the area A on the right panel change. Expand the expression for the area A and write it as a function of x. ^{ 2} + 200xwe might consider the domain of function A(x) as being all values of x in the closed interval [0 , 200] since x >= 0 and y = 200 - x >= 0 (if you solve the second inequality, you obtain x <= 0). To find the value of x that gives an area A maximum, we need to find the first derivative dA/dx (A is a function of x). dA/dx = -2x + 200 If A has a maximum value, it happens at x such that dA/dx = 0. At the endpoints of the domain we have A(0) = 0 and A(200) = 0. dA/dx = -2x + 200 = 0 Solve the above equation for x. x = 100 dA/dx has one zero at x = 100. The second derivative d ^{ 2}A/dx^{ 2} = -2 is negative. (see calculus theorem on using the first and second derivative to determine extremma of functions). The value of the area A at x = 100 is equal to 10000 mm^{ 2} and it is the largest (maximum). So if you select a rectangle of width x = 100 mm and length y = 200 - x = 200 - 100 = 100 mm (it is a square!), you obtain a rectangle with maximum area equal to 10000 mm^{ 2}.
Exercises
1 - Solve the same problem as above but with the perimeter equal to 500 mm. solution to the above exercise
width x = 125 mm and length y = 125 mm. More references on calculus problems |