A rectangle has a perimeter of 40 meters. Find the dimensions that maximize its area.
Solution
Problem
Step 1: Define the variables.
Let the length be x meters and the width be y meters.
The perimeter is 40 meters, so:
2x + 2y = 40
Divide by 2:
x + y = 20
Solve for y:
y = 20 - x
Step 2: Write the area as a function of one variable.
The area of a rectangle is:
A = xy
Substitute y = 20 - x:
A(x) = x(20 - x)
A(x) = 20x - x^2
Step 3: Differentiate the area function.
A'(x) = 20 - 2x
Step 4: Find the critical point.
Set the derivative equal to zero:
20 - 2x = 0
2x = 20
x = 10
Step 5: Find the corresponding width.
y = 20 - x
y = 20 - 10
y = 10
Step 6: Verify that this gives a maximum.
The second derivative is:
A''(x) = -2
Because A''(x) is negative, the area function is concave down, so the critical point gives a maximum.
Therefore, the rectangle with maximum area has dimensions 10 meters by 10 meters.
Result
10 m by 10 m