A gutter with trapezoidal cross section is to be made from a long sheet of tin 8 in wide by turning up one third of its width on each side , what width a cross the top that will give a maximum capacity
Let the sides be turned up through an angle θ. Then the depth of the gutter will be 8/3 sinθ, and the cross-section of the gutter will have area
a = ((8/3) + 2*(8/3)cosθ)/2 * (8/3)sinθ
= 32/9 (sinθ + sin2θ)
for maximum area, we want da/dθ = 0, so
cosθ + 2cos2θ = 0
4cos^2θ + cosθ - 2 = 0
cosθ = 0.593
So, the width across the top for maximum area = 8/3 (1+2*0.593) = 5.83
Hmm. I was expecting the rectangular portion of the trapezoid to be a square, but that turns out not to be so. Oh, well. Still, check my math...
Oops. I messed up the length of the top base. It is 8/3 + 2*(8/3)cosθ
So, the area is
((8/3) + 8/3 + 2*(8/3)cosθ)/2 * (8/3)sinθ
= 32/9 (2sinθ + sin2θ)
For max area we want
cosθ + cos2θ = 0
2cos^2θ + cosθ - 1 = 0
(2cosθ-1)(cosθ+1) = 0
cosθ = 1/2
So, the top base is 8/3 + 2(8/3)(1/2) = 16/3.
Aha! The rectangular portion of the trapezoid is indeed a square!
To find the width across the top that will give a maximum capacity for the gutter, we need to maximize the area of the cross-section.
Let's assume the width across the top of the gutter is "x" inches.
Given that the sheet of tin is 8 inches wide and one-third of its width is turned up on each side, the base of the trapezoidal cross-section will be (8 - x/3 - x/3) inches.
The height of the trapezoid is 8 inches, which is the same as the width of the sheet.
The formula for the area of a trapezoid is (1/2) * (sum of the bases) * height.
So, the area of the cross-section, A, is given by:
A = (1/2) * (8 + (8 - x/3 - x/3)) * 8
Simplify the equation:
A = (1/2) * (16 - x/3 - x/3) * 8
A = (1/2) * (16 - 2x/3) * 8
A = (64 - 8x/3) square inches
To find the width across the top that will maximize the area, we need to find the value of x that maximizes A. This can be done by finding the maximum or minimum of the function.
Differentiate A with respect to x using the chain rule:
dA/dx = (-8/3)
Set dA/dx = 0 to find the critical point:
(-8/3) = 0
Since this equation has no solution, it means that there is no critical point.
To verify that the area is indeed maximized, we can check the limits of the function as x approaches positive infinity and negative infinity.
As x approaches positive infinity, A also approaches positive infinity.
As x approaches negative infinity, A approaches positive infinity.
Since A approaches infinity in both directions, it means that the area is maximized when x is either positive infinity or negative infinity.
Therefore, there is no width across the top that will give a maximum capacity for the gutter.
To find the width across the top that will give a maximum capacity for the gutter, we need to apply some calculus concepts.
Let's denote the width across the top of the gutter as "x". Since we are turning up one third of the width on each side, the width of the bottom of the gutter will be (8 - (1/3)x). The height of the gutter can be denoted as "h".
To determine the capacity of the gutter, we need to find the area of the cross-section. The cross-section of the gutter can be approximated as a rectangle on top, two triangles on the sides, and a rectangle on the bottom.
The area A of the cross-section is given by:
A = (1/2)h((8 - (1/3)x) + x)
Now, we need to find the maximum value of A with respect to x. To do that, we differentiate A with respect to x and set it equal to zero:
dA/dx = (1/2)h(1/3 - 1/2) = 0
1/3 - 1/2 = 0
1/6 = 0
Since there is no solution to the equation, the maximum value of A occurs at the endpoints of the domain, which are x = 0 (when there is no gutter) and x = 8 (where the width of the gutter is equal to the width of the original sheet of tin).
Now, we can evaluate the capacity at x = 0 and x = 8 to determine the maximum capacity.
When x = 0, the width of the gutter is zero, so the capacity is zero.
When x = 8, the width of the gutter is equal to the width of the original sheet of tin, which means the gutter covers the entire sheet. In this case, the capacity is maximized.
Therefore, the width across the top that gives a maximum capacity is 8 inches, which covers the entire width of the original sheet of tin.