A simple rain gutter is constructed using a sheet of aluminum that is 40cm wide. The edges are turned up to form right angles. Determine the depth of the gutter that will maximize the cross-sectional area (allowing the greatest amount of water to flow).
I don't know where and how to start.
please help and thank you
make a sketch.
Assuming that the gutter is rectangular and the sides must have the same height, look at the cross section of the gutter.
let the base by y and each of the sides by x
we know that
2x + y = 40
y = 40-2x
the area A of the cross-section will be
A = xy
= x(40-2x) = 40x - 2x^2
Assuming this is a typical Calculus question ...
dA/dx = 40 - 4x = 0 for a max of A
40-4x = 0
x = 10
the the depth will be 10 cm
if you don't know Calculus , you will have to complete the square on
A = 40x - 2x^2
I don't know calculus
This is what I got after completing the square:
A=-2(x-10)^2+200
How do I know what is the depth when I complete the square?
thank you
To determine the depth of the gutter that will maximize the cross-sectional area, we can use calculus. Here's a step-by-step solution:
Step 1: Draw a diagram or visualize the gutter's cross-section. Let's assume the gutter has a rectangular cross-section, consisting of a base and two opposite sides that are turned up to form a right angle.
Step 2: Label the dimensions. Let's say the base of the gutter is 'x' centimeters and the height (depth) is 'h' centimeters.
Step 3: Write an equation for the cross-sectional area of the gutter in terms of 'x' and 'h'. The area of a rectangle is given by the formula: Area = length × width. In this case, the length is 'x' and the width is the total width of the aluminum sheet (40 cm), which is divided into two sides that are turned up. Thus, the width is 40 - 2x. Therefore, the area A of the gutter is: A = x(40 - 2x) = 40x - 2x^2.
Step 4: To find the depth that maximizes the area, we need to find the value of 'h' that makes the derivative of the area equation equal to zero. Differentiate the area equation with respect to 'x', and set it equal to zero: dA/dx = 0.
Step 5: Calculate dA/dx. Taking the derivative of 40x - 2x^2 with respect to 'x' gives: dA/dx = 40 - 4x.
Step 6: Set the derivative equal to zero and solve for 'x'. 40 - 4x = 0 ⇒ 4x = 40 ⇒ x = 10.
Step 7: Substitute 'x = 10' back into the equation for the area to find the corresponding depth 'h'. A = 10(40 - 2(10)) = 10(40 - 20) = 10(20) = 200.
Therefore, the depth of the gutter that maximizes the cross-sectional area is 200 cm.
To determine the depth of the gutter that maximizes the cross-sectional area, we can use calculus. Let's break down the problem into steps:
Step 1: Define the variables:
Let the depth of the gutter be "x" (in cm).
Step 2: Determine the expression for the cross-sectional area:
The cross-sectional area of the gutter can be calculated by multiplying the depth by the width of the aluminum sheet.
Cross-sectional area (A) = x * 40 cm
Step 3: Calculate the derivative of the cross-sectional area:
Take the derivative of the cross-sectional area function A with respect to x. This step is necessary because we need to find the maximum value of A, which occurs when the derivative equals zero.
dA/dx = 40 cm (derivative of x with respect to x is 1)
Step 4: Set the derivative equal to zero and solve for x:
Setting dA/dx = 0, we have:
40 = 0
Since this equation cannot be satisfied, it means that the derivative does not exist at this point. Therefore, we need to investigate the endpoints of the possible range of x.
Step 5: Consider the endpoints:
Since the gutter is constructed from a sheet of aluminum that is 40 cm wide, the maximum possible value for the depth is 20 cm (half of the width).
Step 6: Evaluate the cross-sectional area at both endpoints:
Substitute x = 0 and x = 20 cm into the expression for the cross-sectional area (A = x * 40 cm) to evaluate the area function at the endpoints:
A(0) = 0 * 40 cm = 0 cm² (When x = 0, the area is 0)
A(20) = 20 cm * 40 cm = 800 cm² (When x = 20, the area is 800)
Step 7: Determine the depth that maximizes the cross-sectional area:
Comparing the areas at the endpoints, we can see that A(20) = 800 cm², which is the maximum area possible. Therefore, the depth of the gutter that maximizes the cross-sectional area and allows the greatest amount of water to flow is x = 20 cm.
In conclusion, the depth of the gutter that will maximize the cross-sectional area and allow the greatest amount of water to flow is 20 cm.