Question
A rain gutter is made from sheets of aluminum that are 28 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 75 square inches. Show that there are two different solutions to the problem. Round to the nearest tenth of an inch.
28
28
Question content area bottom
Part 1
The depth of the gutter is
enter your response here inches.
28
28
Question content area bottom
Part 1
The depth of the gutter is
enter your response here inches.
Answers
Answered by
GPT-4o mini
To solve for the depth of the rain gutter that will allow for a cross-sectional area of 75 square inches, we start by defining a variable for the depth of the gutter. Let \( h \) be the depth of the gutter in inches.
Since the rain gutter is formed from a 28 inch wide aluminum sheet and the edges are turned up at right angles, the width of the bottom part of the gutter will be \( (28 - 2h) \) inches (this is because the depth \( h \) is turned up on both sides).
The cross-sectional area \( A \) of the gutter can be expressed as:
\[
A = \text{width} \times \text{depth}
\]
\[
A = (28 - 2h)h
\]
We want the cross-sectional area to be 75 square inches, so we set up the equation:
\[
(28 - 2h)h = 75
\]
Expanding this equation, we obtain:
\[
28h - 2h^2 = 75
\]
Rearranging it gives:
\[
2h^2 - 28h + 75 = 0
\]
Now, we can apply the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In this case, \( a = 2 \), \( b = -28 \), and \( c = 75 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-28)^2 - 4(2)(75) = 784 - 600 = 184
\]
Now, applying the quadratic formula:
\[
h = \frac{-(-28) \pm \sqrt{184}}{2(2)}
\]
\[
h = \frac{28 \pm \sqrt{184}}{4}
\]
Calculating \( \sqrt{184} \):
\[
\sqrt{184} \approx 13.6
\]
Substituting this value back, we have:
\[
h = \frac{28 \pm 13.6}{4}
\]
Calculating the two possible values for \( h \):
1. For the positive root:
\[
h_1 = \frac{28 + 13.6}{4} = \frac{41.6}{4} \approx 10.4
\]
2. For the negative root:
\[
h_2 = \frac{28 - 13.6}{4} = \frac{14.4}{4} \approx 3.6
\]
Thus, the depths of the gutter that will allow a cross-sectional area of 75 square inches are approximately:
- \( h_1 \approx 10.4 \) inches
- \( h_2 \approx 3.6 \) inches
Therefore, the two different solutions to the problem, rounded to the nearest tenth of an inch, are \( 10.4 \) inches and \( 3.6 \) inches.
Since the rain gutter is formed from a 28 inch wide aluminum sheet and the edges are turned up at right angles, the width of the bottom part of the gutter will be \( (28 - 2h) \) inches (this is because the depth \( h \) is turned up on both sides).
The cross-sectional area \( A \) of the gutter can be expressed as:
\[
A = \text{width} \times \text{depth}
\]
\[
A = (28 - 2h)h
\]
We want the cross-sectional area to be 75 square inches, so we set up the equation:
\[
(28 - 2h)h = 75
\]
Expanding this equation, we obtain:
\[
28h - 2h^2 = 75
\]
Rearranging it gives:
\[
2h^2 - 28h + 75 = 0
\]
Now, we can apply the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In this case, \( a = 2 \), \( b = -28 \), and \( c = 75 \).
Calculating the discriminant:
\[
b^2 - 4ac = (-28)^2 - 4(2)(75) = 784 - 600 = 184
\]
Now, applying the quadratic formula:
\[
h = \frac{-(-28) \pm \sqrt{184}}{2(2)}
\]
\[
h = \frac{28 \pm \sqrt{184}}{4}
\]
Calculating \( \sqrt{184} \):
\[
\sqrt{184} \approx 13.6
\]
Substituting this value back, we have:
\[
h = \frac{28 \pm 13.6}{4}
\]
Calculating the two possible values for \( h \):
1. For the positive root:
\[
h_1 = \frac{28 + 13.6}{4} = \frac{41.6}{4} \approx 10.4
\]
2. For the negative root:
\[
h_2 = \frac{28 - 13.6}{4} = \frac{14.4}{4} \approx 3.6
\]
Thus, the depths of the gutter that will allow a cross-sectional area of 75 square inches are approximately:
- \( h_1 \approx 10.4 \) inches
- \( h_2 \approx 3.6 \) inches
Therefore, the two different solutions to the problem, rounded to the nearest tenth of an inch, are \( 10.4 \) inches and \( 3.6 \) inches.