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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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.

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

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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))`


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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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.

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