.
Since one side is the river, the rectangle's fence perimeter will be
L + 2W = 80.
Hence, L = 80 - 2W.
Area = Length * Width.
Substitute (80-2W) for L:
A = W(80 - 2W)
A = -2W^2 + 80W.
This is a quadratic function. It has the maximum at x = -b/(2a), according to the general theory.
(See the lessons
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
in this site).
For our quadratic function the maximum is at
W = = = 20.
So, W = 20 meters is the width for max area.
Then the length is L = 80 - 2W = 80 - 2*20 = 40 meters
Find the max area. Substitute 20 for W
A = -2(20^2) + 80*20 = 800 square meters.
The plot of the quadratic function for the area is shown below: y = area and x = width.
My other lessons in this site on finding the maximum/minimum of a quadratic function are
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
- HOW TO complete the square to find the vertex of a parabola
- Briefly on finding the vertex of a parabola
- A rectangle with a given perimeter which has the maximal area is a square
- A farmer planning to fence a rectangular garden to enclose the maximal area
- A rancher planning to fence two adjacent rectangular corrals to enclose the maximal area
- Finding the maximum area of the window of a special form
- Using quadratic functions to solve problems on maximizing revenue/profit
- Minimal distance between sailing ships in a sea
- Advanced lesson on finding minima of (x+1)(x+2)(x+3)(x+4)
- OVERVIEW of lessons on finding the maximum/minimum of a quadratic function
Use this file/link ALGEBRA-I - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-I.