Second Derivatives and Beyond
Introduction to Second Derivatives And Beyond - At A Glance:
There are three steps to solving a math problem.
- Figure out what the problem is asking.
- Solve the problem.
- Check the answer.
Apply these steps to a sample optimization problem .
A 12-foot rope is to be cut into two pieces. One piece will lay across the bottom of a wall hanging, and the other piece will lay along the side of the wall hanging.
Where should the rope be cut to maximize the area of the wall hanging?
1. Figure out what the problem is asking.
The problem starts out with a 12-foot rope, which is to be cut into two pieces:
The pieces will then be rearranged like this to frame the wall hanging:
The problem says "where should the rope be cut to maximize the area of the wall hanging?" There are two things in here we need to translate into math.
The first thing is the phrase "where should the rope be cut." We need a mathematical way to represent this. Use a variable x that measures how many feet from the left end we cut the rope:
The second thing is the phrase "the area of the wall hanging". If we cut the rope x feet from the left end, the remaining piece of rope will be ( 12 - x ) feet long.
The wall hanging will look like this:
It's area will be
A(x) = x( 12 - x ) square feet
We've finally figured out what the problem is asking: find the value of x at which A attains its maximum value.
2. Solve the problem.
Now that we've translated the problem into math, this part isn't too bad. We have a function
A(x) = x( 12 - x ) = 12x – x2.
We need to find where A has a maximum. We start by taking the derivative:
A'(x) = 12 - 2x.
This is zero when x = 6, so that's our only critical point.
Use the second derivative test to see if this critical point is a maximum or a minimum. The second derivative is
A"(x) = -2
which is less than zero, so the function A is concave down. This means A must attain its maximum at the critical point x = 6.
Translating back into English, we want to cut the rope 6 feet from the end, or exactly in half.
P.S. A word about endpoints: In this problem, we know 0 ≤ x ≤ 12, because we can't cut the rope at a point that's not on the rope in the first place.
We can't have x = 0 or x = 12, however. If we did, there won't be any wall hanging because one of its dimensions would be zero.
3. Check the answer.
We want to convince ourselves, using common sense, that we really do find the largest wall hanging if we cut the rope exactly in half. One way to convince ourselves is to calculate the area of the wall hanging when the rope is cut exactly in half, then calculate the area of the wall hanging if we cut the rope in some other places.
When the rope is cut exactly in half, the area of the wall hanging is 36 square feet.
If we cut the rope only 3 feet from the end, the area of the wall hanging would be only 27 square feet.
If we cut the rope only 1 foot from the end, the area of the wall hanging would be 11 square feet.
It's believable that we find the largest hanging when the rope is cut exactly in half.