A 10-foot section of gutter is made from a 12-inch-wide strip of sheet metal by folding up 4-inch strips on each side so that they make the same angle with the bottom of the gutter. Determine the depth of the gutter that has the greatest carrying capacity.

greatest carrying capacity ---> greatest volume.

Since the volume = 120(area of end cross-section)
the greatest volume is a function of the surface area of the end
let the length to be folded up be 4 inches, with an angle of Ø

looking at one of the two equal right-angled triangles,
let the height be h
sinØ = h/4
h = 4sinØ -------> we need that depth !
base of triangle -- b
cosØ = b/4
b = 4cosØ

SA = 2 triangles + rectangel
= 2(1/2)(4cosØ)(4sinØ) + 4sinØ(4)
= 16sinØcosØ+ 16sinØ
= 16sinØ(cosØ + 1)

d(SA)/dØ = 16[ sinØ(-sinØ) + cosØ(cosØ+1)]
= 0
-sin^2Ø +cos^2 Ø + cosØ = 0
cos^2Ø - 1 + cos^2Ø + cosØ = 0
2cos^2Ø + cosØ - 1 = 0
(2cosØ - 1)(cosØ + 1) = 0
cosØ = 1/2 or cosØ = -1, the last one would be a fold of 180°, ---> not very likely

if cosØ = 1/2, Ø = 60°
and h = 4sin60°
= 4(√3/2)
= 2√3 inches or appr 3.464 inches

To determine the depth of the gutter that has the greatest carrying capacity, we need to find the maximum area of the cross-section.

Let's assume the depth of the gutter is "x".
The width of the base of the gutter (the bottom) is 12 inches.
The width of each folded-up strip is 4 inches.

To calculate the area, we need to subtract the width of the base from the total width and divide it into two equal parts (since there are two folded-up strips).

Width of each folded-up strip = (12 inches - 4 inches) / 2 = 4 inches.

Now, we can determine the width of the top of the gutter (the folded-up portion).
Width of the top = 4 inches + 4 inches = 8 inches.

Therefore, the width of the gutter is 8 inches throughout its depth.

The area of the cross-section of the gutter is given by the formula: Area = (width of the base + width of the top) * (depth) / 2.

So, the area of the cross-section is: Area = (12 inches + 8 inches) * (x inches) / 2 = 10x square inches.

In order to maximize the carrying capacity, we need to maximize the area of the cross-section.

Since the length of the section is fixed at 10 feet, we only need to consider the depth (x) of the gutter.

Now, let's take the derivative of the area function with respect to x to find the critical point(s) where the maximum area occurs.

d(Area)/dx = 10

Since the derivative is positive, this means that the area is increasing as the depth of the gutter increases. Since there are no critical points, this means that the area is always increasing, and therefore, the depth of the gutter can be infinite to achieve the maximum carrying capacity.

Therefore, the depth of the gutter that has the greatest carrying capacity can be as deep as desired, as long as the length of the gutter section is 10 feet.

To determine the depth of the gutter that has the greatest carrying capacity, we need to find the maximum area of the cross-section of the gutter.

Let's start by visualizing the situation. We have a 10-foot (120-inch) section of gutter made from a 12-inch-wide strip of sheet metal. The gutter is formed by folding up 4-inch strips on each side, creating a trapezoidal cross-section.

To find the depth of the gutter, we can use the Pythagorean theorem. Let's denote the depth of the gutter as "d". The 4-inch fold on each side forms a right triangle with the bottom of the gutter, where the hypotenuse is the depth "d" and the two legs are 4 inches. Therefore, the equation is:

d^2 = 4^2 + 4^2
d^2 = 16 + 16
d^2 = 32

Taking the square root of both sides, we find that:

d ≈ √32
d ≈ 5.66 inches

So, the approximate depth of the gutter that maximizes the carrying capacity is 5.66 inches.