A farmer has 100 yeards of fencing with which to enclose two adjacent rectangular pens both bordering a river. The farmer does not need to fence the side with the river. What should the dimensions of the two pens together (a rectangle) be in order to yield the largest possible area?

Perimeter = 3w+4L=100

area= 2LW

but W=(100-4L)/3

area=2L (100-4L)/3

take the derivative with respect to L, set to zero, and solve for L, then go back and find W.

Ooops. You are not a calculus student yet.

area=2L(100-4L)/3= 200L/3 -8L^2/3

This is a parabola, with zeroes at L=25, and L=0. So max must be halfway between, L=25/2

if L = 25/2 then W=(100-50)/3 and that is it.

Thank you bobpursley!

To find the dimensions of the two pens that yield the largest possible area, we need to maximize the total area of both pens.

Let's denote the width of one pen as 'w' and the length as 'L'. Since the two pens are adjacent, the combined width will be 2w. The combined length will be L.

We know that the total length of the fence available is 100 yards. Since we don't need to fence the side bordering the river, we can write the equation:

2w + L = 100

To maximize the area, we need to express the area in terms of a single variable. The area of one pen is given by A = w * L. Since we have two pens, the total area (T) is the sum of the areas of both pens:

T = 2w * L

To simplify the expression further, we can solve the first equation for L:

L = 100 - 2w

Substituting this value of L into the area equation:

T = 2w(100 - 2w)

Expanding and rearranging, the total area equation becomes:

T = 200w - 4w^2

Now, to find the dimensions that maximize the area, we need to find the value of w that maximizes the equation T. We can achieve this by taking the derivative of T with respect to w, setting it equal to zero, and solving for w.

dT/dw = 200 - 8w

Setting dT/dw equal to zero:

200 - 8w = 0

8w = 200

w = 25

Now we have the value of w that maximizes the area. To find the value of L, we can substitute w = 25 into the first equation:

2(25) + L = 100

50 + L = 100

L = 50

So, the dimensions of the two pens that yield the largest possible area are:
Width of each pen (w) = 25 yards
Length of each pen (L) = 50 yards