A rain gutter is made from sheets of aluminum that are 11 inches wide. As shown in the figure the edges are turned up to form right angles. Determine the depth of the gutter that will allow a cross-sectional area of 10 in.². Show that there are two different Solutions to the problem
To determine the depth of the gutter that will allow a cross-sectional area of 10 in.², we can start by visualizing the cross-section of the gutter.
In the provided figure, we have a rectangular cross-section with turned-up edges forming right angles. Let's assume the depth of the gutter is represented by "d" inches.
Now, we can calculate the cross-sectional area of the gutter. Since the gutter is rectangular, the area can be calculated by multiplying the width and the height (depth).
Area = Width × Height
Since the width of the gutter is given as 11 inches, and we want the area to be 10 in.², we can write the equation as:
10 = 11 × d
Now, we can solve for "d":
d = 10 / 11
d ≈ 0.909 inches
So, the depth of the gutter that will allow a cross-sectional area of 10 in.² is approximately 0.909 inches.
Now, to show that there are two different solutions to the problem, let's consider the possibility of a negative depth. If we plug in a negative depth value, the equation becomes:
10 = 11 × (-d)
By solving for "d", we get:
-d = 10 / 11
d ≈ -0.909 inches
Since negative values of depth don't make sense in this context, we discard the negative solution. Hence, there is only one valid solution to the problem, which is a positive depth of approximately 0.909 inches.
To determine the depth of the gutter that will allow a cross-sectional area of 10 in², we can set up an equation using the given information.
Let's assume that the depth of the gutter is "x" inches.
The cross-sectional area of the gutter can be calculated by multiplying the width (11 inches) by the depth (x inches). So, we have:
Cross-sectional area = Width * Depth
10 in² = 11 in * x
Now, let's solve this equation to find the value of x.
Dividing both sides of the equation by 11:
10 in² / 11 in = x
0.909 in ≈ x
So, the depth of the gutter that will allow a cross-sectional area of 10 in² is approximately 0.909 inches.
However, since the edges of the gutter are turned up to form right angles, there are two different solutions to the problem. In addition to the previously found solution, there is another solution as the depth could be negative (-0.909 inches). Both solutions will result in a cross-sectional area of 10 in².
x+2y = 11
xy = 10
(11-2y)y = 10
2y^1-11y+10 = 0
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