Thread: Word problem - Second derivative test, minimum length of fence

A developer of an apartment/office complex wants to fence in a rectangular parking lot area of 45000 square meters, and divide it into three parts with two parallel fences both parallel to one side of the rectangle. Each part will have a different parking lot for employees, handicapped parking, and tenants. What is the shortest total length of fence that he can use?

The answer must be found with the second derivative test to show that the length is a minimum.

Make a start at least.
Then we can see where your difficulties lie.

Originally Posted by angie427

A developer of an apartment/office complex wants to fence in a rectangular parking lot area of 45000 square meters, and divide it into three parts with two parallel fences both parallel to one side of the rectangle. Each part will have a different parking lot for employees, handicapped parking, and tenants. What is the shortest total length of fence that he can use?

The answer must be found with the second derivative test to show that the length is a minimum.

Have you at least started by drawing a diagram?

An overall plan for solution: you get to choose the width of the parking lot, so let that be your variable, call it w. Then you calculate the length of fence as a function of w, and set its derivative to zero. Then solve for w.

If you must use the second derivative test to show it's a minimum, then you must. But it is obvious based on the physical situation.

- Hollywood

Yes I've drawn a diagram. I think I know now what to do from there.