A gutter with trapezoidal cross-section is to be made from a long sheet of stainless steel that is 30 cm wide by turning up one-fourth of the width on each side. What width across the top will give the maximum cross sectional area?
So, the sloping sides are 7.5 each, and the bottom is 15cm wide. Thus, if the top has width 2x+15, we have a cross-section area
a = (2x+15 +15)/2 √((15/2)^2-x^2)
= (x+15)/2 √(225-x^2)
da/dx = -(8x^2+60x-225) / (2√225-x^2))
da/dx=0 when x=2.7452
So, the width for maxim area is 20.49cm
The shape of a patio is triangular with a height of 8 ft and base of 17 ft. What is the cost of outdoor carpeting to cover the patio if the carpeting costs $6.50 per ft2?
To find the width across the top that will give the maximum cross-sectional area, we can use calculus and optimization techniques. Here are the step-by-step instructions:
1. Let's assume the width across the top of the trapezoidal gutter is denoted by "x" cm.
- Since the total width is 30 cm and one-fourth of the width is turned up on each side, the base of the trapezoid is (30 - 2 * 1/4 * x) cm.
2. The height of the trapezoidal gutter is the same as the turned-up width, which is 1/4 * x cm.
3. The cross-sectional area (A) of the trapezoidal gutter can be calculated using the formula:
A = (1/2) * (base1 + base2) * height.
- In this case, the base1 is x cm and the base2 is (30 - 2 * 1/4 * x) cm.
- The height is 1/4 * x cm.
4. Simplify the equation for the cross-sectional area:
A = (1/2) * (x + (30 - 2 * 1/4 * x)) * (1/4 * x)
A = (1/2) * (x + (30 - (1/2 * x))) * (1/4 * x)
A = (1/8) * (x + (30 - (1/2 * x))) * x
5. Expand and simplify the equation further:
A = (1/8) * (x + 30 - (1/2) * x) * x
A = (1/8) * (30 + (1/2) * x) * x
A = (3/16) * (30 + (1/2) * x) * x
6. Now, we need to find the value of x that maximizes the cross-sectional area A. To do this, we take the derivative of A with respect to x and set it equal to zero:
dA/dx = (3/16) * (30 + (1/2) * x) + (3/16) * x
0 = (3/16) * (30 + (1/2) * x) + (3/16) * x
7. Simplify the equation:
0 = (3/16) * (30 + (1/2) * x) + (3/16) * x
0 = (3/16) * 30 + (3/16) * (1/2) * x + (3/16) * x
0 = (3/16) * 30 + (3/32) * x + (3/16) * x
0 = (3/16) * 30 + (6/32) * x + (3/16) * x
0 = (3/16) * 30 + (9/32) * x
8. Solve for x:
(9/32) * x = - (3/16) * 30
x = - (3/16) * 30 * (32/9)
x = - 2 * 30 * (2/3)
x = - 2 * 20
x = -40
9. Since the width cannot be negative, we can disregard the negative value of x. So, the width across the top that gives the maximum cross-sectional area is 40 cm.
Therefore, the width across the top of the trapezoidal gutter that gives the maximum cross-sectional area is 40 cm.
To find the width across the top that will give the maximum cross-sectional area, we need to determine the function that represents the area of the trapezoidal cross-section.
Let's assume the width across the top is x cm. Since we are turning up one-fourth of the width on each side, the width of the base of the trapezoid will be (x - 2 * 1/4 * x) = (x - x/2) = x/2. The height of the trapezoid is the width of the turned-up portion, which is 1/4 * x.
The area of a trapezoid is given by the formula: A = (1/2) * (b1 + b2) * h, where b1 and b2 are the lengths of the parallel sides (the bases) and h is the height.
Applying this formula to the trapezoidal cross-section, the area of the cross-section is:
A = (1/2) * (x/2 + x) * (1/4 * x) = (1/2) * (3x/2) * (1/4 * x) = (3/8) * x^2.
To find the width across the top that gives the maximum cross-sectional area, we need to find the value of x that maximizes the function A(x) = (3/8) * x^2.
To do this, we can take the derivative of A(x) with respect to x, set it equal to zero, and solve for x.
dA/dx = (3/8) * 2x
Setting dA/dx = 0:
0 = (3/8) * 2x
0 = 3x
x = 0
The critical point x = 0 does not make sense in this context since a gutter with zero width would not exist. Therefore, we need to consider the endpoints of the possible range of x.
The given width of the stainless steel sheet is 30 cm, so x has a maximum possible value of 30 cm.
So, to maximize the cross-sectional area, we have two possibilities: x = 0 (which is not valid) and x = 30 cm (the maximum possible value).
Therefore, the width across the top of the gutter that will give the maximum cross-sectional area is 30 cm.