Ken Graham has 1200 meters of fence to put around two rectangular yards. If the yards are to be separated by part of the fence, what would be the length and width for maximum area?

Each yard is 150 meters wide and 200 meters tall.

The overall area is 300 meters wide by 200 meters tall.

Step-by-step explanation:

Let's say each yard has a width w and a height h. And let's say the part of the fence that divides them is parallel to the height. The total amount of fencing is therefore:

1200 = 4w + 3h

The total enclosed area is:

A = 2wh

Start by solving for w in the first equation:

1200 - 3h = 4w

w = 300 - 0.75h

Substitute into the second equation:

A = 2 (300 - 0.75h) h

A = 600h - 1.5h²

Take derivative and set to 0:

dA/dh = 600 - 3h

0 = 600 - 3h

h = 200

Solve for w:

w = 300 - 0.75h

w = 150

So each yard is 150 meters wide and 200 meters tall.

The overall area is 300 meters wide by 200 meters tall.