A farmer wants to enclose the largest possible rectangle using a straight river as one boundary and #2400# feet of fencing. What are the dimensions of the rectangle of largest possible area?

2 Answers
Feb 11, 2016

The maximum area is given by a #600 xx 1200# ft rectangle, being half of a square.


Method 1 - Geometry

If there were no river then the maximum area would be given by a square configuration since increasing the length and decreasing the width of a square by the same value #t# will decrease the area by a square with sides of length #t#.

Now add the river on one side. The river acts like a mirror. As we fence in a rectangle on one side of the river, imagine an identical rectangle on the other side of the river. The maximum area is given when the two rectangles form a square, that is when we have a rectangle twice as long as wide.

Hence with #2400# ft of fence, the rectangle has fenced sides #600# ft, #1200# ft and #600# ft.

Method 2 - Algebra and calculus

Suppose the long side of the rectangle has length #t# ft. Then the short sides have length #(2400 - t)/2# ft and the area of the rectangle is:

#a(t) = (2400 - t)/2 * t = -1/2t^2 + 1200t#

Then #d/(dt) a(t) = -t + 1200#, which is zero when #t = 1200#

Since the coefficient of the leading #t^2# term is negative this value of #t# results in a maximum value for the function #a(t)#

So the longest side is #1200# ft and the remaining sides are #600# ft each.

Feb 11, 2016

The dimensions is 600 ft by 1200 ft.


First, set up the equations:
area= xy
perimeter= 2x+ y =2400 (remember, the river serves as a natural border so the fence only has 3 sides with two of them being equivalent because they are opposite sides of a rectangle.)

Use substitution to get one variable.
2x+ y= 2400
y= 2400- 2x
Plug that into the area equation: xy
x(2400- 2x)
distribute the x
2400x- 2#x^2#= area

There are two ways to do this problem: using calculus or using the calculator.

Using calculus, to find the maximum of the area, the derivative of the area equation should equal zero.
area = 2400x- 2#x^2#
#(da)/dt#=2400 -4x
Set it to zero and solve.
2400 -4x = 0
4x= 2400
x =600

solve for y by plugging x into the perimeter equation
2(600) +y = 2400
1200 + y= 2400
y= 1200

Using the calculator, push the Y= button and enter the area equation, 2400x- 2#x^2#. Then hit the GRAPH button.
You might have to adjust the window to see the full parabolic curve. To calculate the maximum of the area, press the 2ND and then the TRACE button. Then hit number 4 to calculate maximum. It will ask you for left bound which means go to the left of the maximum of the curve and hit enter. Then it will ask you right bound, which means to the left of the maximum and hit enter. It will then ask GUESS? which means go in-between the two left and right bound and hit enter.

It will yield (600, 720000) which means x= 600. Plug that into the perimeter equation to get a y value of 1200. Those are the dimensions of the fence to maximize area.