Maximum Area: a rancher has 200 feet of fencing to enclose two adjacent rectangular corrals. what dimensions should be used so that the enclosed area will be a maximum?

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ishpiro eNotes educator| Certified Educator

Denote the length of the shared side of the corrals by `l` and the width of the both corrals together as `w` (please see the attached image). Then, the required amount of fencing will be 

`P = l + l + l + w + w = 3l + 2w=200 ft` .

The enclosed area then will be `A = w*l` . Expressing area as a function of `l` , we get:

`A = (200-3l)/2 l = (200l - 3l^2)/2 = 100l - 3/2l^2` .

This is a quadratic function and its graph is a parabola. In standard form the equation of a parabola is

`y = ax^2 + bx + c`

If `a <0` , the parabola is upside down and the function then has a maximum value which is achieved at the vertex. The coordinates of the vertex are

`x = -b/(2a)` , `y = y(-b/(2a))` .

Area as a function of the length can be written in standard form as

`A = -3/2l^2 + 100l` .

The leading coefficient is `a = -3/2` , which is negative, so the maximum of this function is at the vertex

`l = -100/(2*(-3/2))`

`l=100/3`

Thus, the area is maximized when the shared side of the corrals has the length of `100/3 = 33 1/3` ft and the total width is then `w = 1/2(200-3l) = 1/2(200 - 3*100/3) = 50` ft. The maximum enclosed area is `A = wl = 5000/3 = 1666 2/3 ` square ft.

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justaguide eNotes educator| Certified Educator

Two adjacent rectangular corrals are to be enclosed with 200 feet of fencing. Let us assume that the dimensions of the two rectangles are the same.

If the dimensions are w and l, the total perimeter of the two rectangular corrals is 4w+4l but with a common side this is equal to 4w + 3l. 

Now the enclosed area is 2*w*l and as the length of the fencing is 200 ft, we have l + 2w + 2w + l + l = 200

3l + 4w = 200

`l = (200 - 4w)/3`

To maximize the enclosed area, we need to maximize: `A = 2*w*l = 2*w*(200 - 4w)/3`

`A = (400w - 8w^2)/3`

Using calculus, equate A' = 0 and solve for w:

`A' = ((400w - 8w^2)/3)'`

`= 400/3 - (16/3)*w`

`400/3 - (16/3)*w = 0`

`=> 400 - 16w = 0`

`=> w = 400/16`

`=> w = 25`

`l = (200 - 4w)/3 = 100/3`

The maximum enclosed area is 5000/3 square feet.

lemjay eNotes educator| Certified Educator

Let x and y be the length and width of the rectangular corral.

Since the two rectangular corrals are adjacent to each other, let one side of each corral (x) share a common fence.So, their total perimeter is:

`P=3x+4y`

Plug-in the given perimeter of the fence.

`200=3x+4y`

Then, isolate either of the variable. Let the y be isolated.

`(200-3x)/4 =y`        (Let this be EQ1).

Next, set-up the equation for total area of the two rectangular corrals.

`A=2(xy)`

Then, express the right side as one variable. To do so, substitute EQ1.


`A=2x(200-3x)/4`

`A=x(200-3x)/2`

`A=(200x-3x^2)/2`

To determine the dimensions of each rectangular corral that would  maximize area, take the derivative of area.

`A'= (200-6x)/2`

`A'=(2(100-3x))/2`

`A'= 100-3x`

Then, set A' equal to zero and solve for x.

`0=100-3x`

`3x=100`

`x=100/3`

Next, substitute the value of x to EQ1.

`y=(200-3x)/4=(200-3(100/3))/4=(200-100)/4`

`y=100/4=25`

Hence, the length and width of each rectangular corrals is `100/3` ft. and 25 ft, respectively.

 

Here is a similar problem with 400 total feet of fencing instead of 200.