THE ULTIMATE RECTANGLE PROBLEM
Suppose that a fixed amount of fencing "f" is available. The goal is to create the rectangle with the
maximum area, under the given conditions. The length of the rectangle will be called "x", and the width "y".
The large rectangle may have some smaller rectangles created within it. If so, all vertical fences are parallel
and all horizontal fences are parallel. All interior fences go completely across the large rectangle. The
widths of interior rectangles do not have to be equal; neither do the lengths.
The number of lengths of fencing required is "a"; "a" will be one or more. The number of widths of fencing
required is "b"; "b" will be one or more.
Examples:
existing barn neighbor’s
fence
river
existing
fence
a = 1, b = 2 a = 1, b = 1 a = 2, b = 1 a = 4, b = 4 a = 6, b = 3
Example: Alex has 1000 feet of fencing with which to enclose garden
plots as illustrated. Determine the length and width of the
outside dimensions so that the total area enclosed is
maximized.
Step 1: Choose a variable, identify it completely; identify other unknowns in terms of same variable.
Let x = length of fenced area
Then = width of fenced area
1000 2
6
−
x
Step 2: Write an equation in terms of your chosen variable which summarizes the given information.
The number of lengths used is 2, i.e. a = 2
The number of widths used is 6, i.e. b =6
Since the goal of the problem is to maximize area, the equation is based on the area of the overall
rectangle : Ax x x x x
( )
=−
=−
1000 2
6
1000 2
6
2
Step 3: Solve your equation.
In this problem, you need to maximize the area. Since the area function is quadratic with a negative
leading coefficient, an absolute maximum exists.
and xb
a
= − = − −=
2
10006
23
250 1000 2 250
683 1
3
−=
( )
Step 4: Answer the question(s) with complete sentences, including units.
The overall rectangle should be 250 ft by ft, with each interior fence being ft long. The total
83 1
383 1
3
area enclosed will be .
20833 1
32
, ft
