Use Derivatives to solve problems: Area Optimization
A problem to maximize (optimization) the area of a rectangle with a constant perimeter is presented. An interactive applet 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.
Problem : You decide to construct a rectangle of perimeter 400 mm and maximum area. Find the length and the width of the rectangle.
Interactive Tutorial
Let us try to understand the problem using the applet below. There are many ways you can construct a rectangle of perimeter 400 mm. But how to obtain one with maximum area.
APPLET
On the left panel of the applet, use the mousse to press and drag point C to increase or decrease the width x of the rectangle. Note that as you increase the width, the length decreases. It is because the perimeter has to stay constant at 400 mm. On the right panel you have the area plotted against x the width of the rectangle. As you can see there are many ways we can select x but there seem to be one value of x for which the area is largest (maximum). Approximate this value of x from the graph. You may also plot the whole graph using the "on" and "off" buttons above it.
Analytical Tutorial
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:
A = x*y
The perimeter may be written as
P = 400 = 2x + 2y
Solve equation 400 = 2x + 2y for y
y = 200  x
We now now substitute y = 200  x into the area A = x*y to obtain .
A = x*(200  x)
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.
A(x) = x^{ 2} + 200x
we 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 
