A pig rancher wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. He has 610 feet of fencing available to complete the job. What is the largest possible total area of the four pens?

two lengths and five widths ... 2 L + 5 W = 610 ... L = 305 - 2.5 W

A = L * W = 305 W - 2.5 W^2

the maximum is on the axis of symmetry of the parabola

Wmax = -b / 2a = -305 / -5 = 61

substitute back to find L ... then calculate the area

Well, I guess those 4 pens are to be the same size.

He could do it with 4 fences parallel with the wall
or
he could do it with 5 fences perpendicular to the wall.
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If he does it with 4 plus two ends:
A = 4 x y with y parallel to wall and length 4 x perpendicular to the wall
then
610 = 4 y + 8 x
4 y = 610 - 8 x
y = (610/4) - 2 x
then
A = 4 x [ (610/4) - 2 x]
A = 610 x - 8 x^2
that is min when dA/dx = 0 (or find vertex of that parabola)
dA/dx = 0 = 610 - 16 x
x = 610 /16 = 305/8 or about 38.125
then y = (610/4) - 2 x = (610/4) - 610/8 = 610/8
then Area = 4 x y = 4 (305/8)(610/8)= 305*610/16 = 11,628.125
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BUT the other way, 5 x lengths and one y parallel to wall
610 = 5x + y
A = x y
then A = x (610-5x) = 610 x - 5 x^2
dA/dx =0 = 610 - 10 x
x = 61
then y = 610 - 5(61) = 305
and
x y = Area = 61*305 = 18,605 THE WINNER

Scott - the wall so only one length

Damon ... don't see "wall" mentioned in the problem ...

There will be 2 ends and 5 cross fences. Maximum area is achieved when the fencing is divided equally among lengths and widths: 305 feet each.

So, the dimensions will be 152.5 by 61

You are right Scott. I misread it.

To find the largest possible total area of the four pens, we need to determine the dimensions of the rectangle that would maximize the area.

Let's assume the length of the rectangle is L, and the width is W. Since we want to divide the rectangle into four pens with fencing parallel to one side, two sides will be used as fencing, and two sides will be shared between the adjacent pens.

The total length of the fencing required for the rectangle is twice the length plus twice the width, which can be expressed as:

2L + 2W

According to the given information, the pig rancher has 610 feet of fencing available. Therefore, we have:

2L + 2W = 610

Now, we need to express the area of the rectangle in terms of L and W. The area of the rectangle is simply the product of its length and width:

Area = L * W

To find the largest possible total area, we need to maximize this function while satisfying the fencing constraint.

We can solve for one of the variables in terms of the other using the fencing constraint equation:

2L + 2W = 610

Simplifying the equation:

L + W = 305

Solving for L:

L = 305 - W

Now substitute this expression for L in the area equation:

Area = (305 - W) * W

Expand the equation:

Area = 305W - W^2

To maximize the area, we can take the derivative of the area function with respect to W and set it equal to zero:

d(Area)/dW = 305 - 2W = 0

Solving for W:

305 - 2W = 0
2W = 305
W = 305/2 = 152.5

Now substitute the value of W back into the expression for L:

L = 305 - W = 305 - 152.5 = 152.5

So, the dimensions of the rectangle that would maximize the total area are L = 152.5 and W = 152.5.

To find the largest possible total area, substitute these values into the area equation:

Area = L * W = 152.5 * 152.5 = 23256.25 square feet

Therefore, the largest possible total area of the four pens is 23256.25 square feet.