A long rectangular sheet of metal, 12 inches wide is to made into a rain gutter by turning up two sides at angles of 120° to the sheet. How many inches should be turned up to give the gutter its greatest capacity?

If the sides of the gutter have length x, then the maximum capacity happens when the cross-section has maximum area. That is a trapezoid.

The depth of the gutter is x sin60°
So, the trapezoid has area a = (12-2x + 2x cos60°)/2 * x sin60°
maximum area is where da/dx = 0
Now take it away. I get x=4

oops

a = (12-2x + 12-2x+2x cos60°)/2 * x sin60°

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Socratic › ... › Solving Optimization Problems
A long rectangular sheet of metal, 12cm wide, is to be made ...

The result is in cm not in inches but it doesn't matter.

Remark:

It is written:
y' = ( 12 − 3 x ) ( √3 / 2 ) = 0

It should be written:
A' = ( 12 − 3 x ) ( √3 / 2 ) = 0

Now multiply both sides by 2 / √3

12 - 3 x = 0

12 = 3x

12 / 3 = x

4 = x

x = 4

Second derivative test:

If A" ( x ) < 0 then function has a local maximum at x.

If A" ( x ) > 0 then function has a local minimum at x.

A" ( x ) = - 3 ∙ √3 / 2 = - 2.598 < 0

Function has a local maximum at x = 4

To find the length that should be turned up to give the gutter its greatest capacity, we need to calculate the area of the gutter for different lengths turned up and determine which length gives the maximum area.

Let's denote the length to be turned up as "x" inches. Since we need to turn up two sides at angles of 120° to the sheet, we will have a triangular cross-section for the gutter. The base of the triangular cross-section will be the width of the sheet, which is 12 inches.

To find the height of the triangular cross-section, we can use trigonometry. In a triangle with angles of 120°, 30°, and 30°, the side opposite the 30° angle is half the length of the hypotenuse.

In our case, the hypotenuse is the length turned up, which is "x" inches. Therefore, the height of the triangular cross-section is (1/2)x inches.

Now, we can calculate the area of the triangular cross-section using the formula for the area of a triangle (A = 1/2 * base * height). The base is 12 inches, and the height is (1/2)x inches.

So, the area of the triangular cross-section is A = 1/2 * 12 * (1/2)x = 6x/4 = 3x/2 square inches.

To maximize the capacity of the gutter, we want to maximize the area. Hence, we need to determine the value of "x" that maximizes the area.

The maximum value of "x" can be found by taking the derivative of the area function with respect to "x" and setting it equal to zero. However, in this case, since the area function is a linear function (A = 3x/2), there is no critical point. Hence, the area of the triangular cross-section will continue to increase as "x" increases.

Therefore, to maximize the area (and thus the capacity) of the gutter, we would turn up the entire length of the sheet. Thus, "x" should be equal to the length of the sheet, which is not given in the question.