A 3D visual representation of a long rectangular sheet of metal, 12 inches wide, in the process of being fashioned into a rain gutter. Two sides of the sheet are being turned up, forming a right angle with the sheet and transforming it into a three-sided trough. Keep the dimensions visible but refrain from incorporating any numerals or alphabetic text. Include a few droplets of rain falling into the gradually forming gutter to emphasize its purpose. Use shades of grey and silver to depict the metal sheet and gutter, in contrast with the vibrant blue of the rain droplets.

A long rectangular sheet of metal, 12 inches wide, is to be made into a rain gutter by turning up two sides so that they are perpendicular to the sheet. How many inches should be turned up to give the gutter its greatest capacity?

All we have to consider is the area of the cross-section, since the length is a constant.

let the sides to be turned up at both ends be x inches long
let the remaining base be y inches
2x + y = 12
y = 12-2x

area = xy = x(12-2x) = 12x - 2x^2
d(area)/dx = 12 - 4x = 0 for a max of area
4x = 12
x = 3

So the turn-ups should be 3 inches

7inches long 5 inches wide

Well, I'm glad you asked! To find the greatest capacity of the rain gutter, you'll need to turn up the sides by an equal amount. Let's call that amount "x".

When both sides are turned up by "x" inches, the width of the gutter will be reduced by 2x inches. Therefore, the width of the gutter would be 12 - 2x inches.

Now, the capacity of a rectangular gutter is determined by its depth and width. The depth is already fixed (since it remains the same as the original sheet), so we only need to find the width.

To maximize the width and subsequently, the capacity, we need to find the value of x that makes the width as large as possible.

Since the width is 12 - 2x, we can see that it will be the largest when x is as small as possible. In this case, that's when x is 0. Turning up the sides by 0 inches would give us the maximum width of 12 inches.

So, to give the rain gutter its greatest capacity, you should not turn up any inches. Just leave it as it is and let the rain flow freely!

To find the length of the sides that need to be turned up to give the gutter its greatest capacity, we can use the concept of optimization.

Let's assume that the length of the rectangular sheet of metal is 'x' inches. We need to find the value of 'x' that maximizes the capacity of the gutter.

For a rectangular gutter, the cross-section will be in the shape of an isosceles trapezoid. The area of an isosceles trapezoid can be calculated using the formula:

Area = (1/2) * (a + b) * h

where 'a' and 'b' are the lengths of the parallel sides and 'h' is the height (the length of the turned-up sides).

In this case, the width of the sheet (12 inches) will be the sum of the lengths of the parallel sides, which is a + b. Therefore, we can rewrite the formula as:

Area = 6x * h

We want to maximize the area, so we need to find the value of 'h' that gives the maximum area.

To do that, let's differentiate the formula for the area with respect to 'h' and solve for the value of 'h' when the derivative is equal to zero.

d(Area)/dh = 6x

Setting the derivative equal to zero:

6x = 0

Solving for 'x':

x = 0

Since 'x' cannot be zero (as it represents the length of the sheet), we conclude that there is no maximum area. This implies that turning up the sides of any length will give the gutter its greatest capacity. The length of the sides to be turned up can be any value.

Greatest capacity will be achieved when the area of cross section will be maximum.

Let X be the base ,so (12-X)/2 will be the height.
Area= Base x height= X(12-X)/2
=(12X-X^2)/2
Following the maximum theory, we differentiate the area w.r.t X and equate it to zero.
d/dX of(Area)=(12-2X)/2 =0
or 6-X=0
or X=6 inches(base)
so, height= (12-6)/2 = 3 inches

x (2x-2)

d/dX of(Area)=(12-2X)/2 =0
or 6-X=0
or X=6 inches(base)
so, height= (12-6)/2 = 3 inches