Question

A piece of rectangular sheet metal is 20 inches wide. Its is to be made into a rain gutter by turning up the edges to form parallel sides. let x represent the length of each of the parallel sides. a) Give the restrictions on x b) Determine a function A that gives the area of the cross section of the gutter. C) for what value of x will A be a maximum (and thus maximize the amount of water that the gutter will hold) What is the max area?

Answer 1

can you add figure??

Answer 2

Turning up the edges of the sheet metal will result in the 20 inches of metal being divided into x inches of the first side, then the unknown width of the gutter (let's call it "y") and then again x inches for the second side.

As a picture:
x|______|x
y

As a formula:
20 = x + y + x = 2x + y

Resolving this to y we get:
20 - 2x = y
and switching it around:
y = 20 - 2x, I think this should reply part a) of your question.

I'll continue in a second post to let you digest this first and see if you agree...

Answer 3

Well, correcting myself: The question asks us to give the restrictions on x, I therefore should probably have resolved to x rather than to y:
2x = 20-y
x = (20-y)/2

Answer 4

Now for part b) of your question: The area of the gutter's cross-section is its width (y) multiplied with its height (x).

A(x,y) = x * y

If we use our result from a) to eliminate y then we can see that
A(x) = x * y = x * (20-2x) = 20x-2x^2

So I'd say that the answer to part b) should be:
A(x) = 20x - 2x^2

Answer 5

And finally part c) of the question. I could have resolved the formula for the cross section to y as well in the last post, getting A(y). But I resolved it to x so that we can answer part c) which asks to find the x that maximizes the area. So we need a function of x to achieve that: A(x).

Now to find the x that maximizes A(x) it's probably a good idea that you first try to picture A(x) as a graph. For very large and very small x, the negative 2x^2 will be much more important than the positive 20x. This means that you function will start very small then go up to a maximum and then go down again. I attached a graph of the function to make this clear.

So what you want is to find the maximum which is the top turning point of your function. You get this by taking the first derivative of the function and then finding the places where this derivative is null. So far so good?

Answer 6

Now the first derivative of 20x-2x² is 20-4x. And this derivative is zero exactly at the turning point of the function.

So let's solve the equation 20-4x = 0 to find that turning point:

20-4x = 0
-4x = -20
4x = 20
x = 20/4
x = 5

So x = 5 maximizes the area A(x) of the cross-section of the gutter.

To answer the final part of the question we just have to find the y corresponding to x = 5. We can use the formula for y that we noted down at the beginning:

y = 20 - 2 * x

In our case then:

y = 20 - 2 * 5 = 10

The area of the gutter's cross-section for x=5 and y=10 is

A(x,y) = x*y = 50

(in squared inches).

So the maximum area of the gutter's cross-section is 50 squared inches.

We could have gotten the same result a bit less intuitively inserting x=5 directly in our function A(x) = 20x - 2x² = 20 * 5 - 2 * 5² = 100 - 2 * 25 = 100 - 50 = 50. This is a nice way to check that you (I mean: I) didn't make a mistake earlier.

Hope this helps!