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A rain gutter is made from sheets of aluminum that are 24 in wide. The edges are turned up to form right angles. Determine the depth of the gutter that will allow a cross-sectional area of 52 sq in.

2007-12-09 12:53:44 · 2 answers · asked by Boonay444 2 in Science & Mathematics Mathematics

2 answers

To make the gutter, two sides will need to be bent
upward from the bottom. Call the size of each side
side "y" and the size of the bottom "x". Thus
we know that:

bottom + left side + right side = 24
x + y + y = 24
x + 2y = 24

Solving for y:
x + 2y = 24
2y = 24 - x
y = (1/2)(24 - x)
y = 12 - (1/2)x

We know that the cross sectional area is just the
bottom dimension times the side dimension:

x * y = 52
x * (12 - (1/2)x) = 52
12x - (1/2)x^2 = 52
0 = (1/2)x^2 - 12x + 52
0 = x^2 - 24x + 104

Using the quadratic equation:

x = [-b +/- sqrt(b^2 - 4ac)]/2a
x = [24 +/- sqrt(24^2 - 4(1)(104)]/2
x = [24 +/- sqrt(160)]/2

x = 5.68 y = 9.16

To check:
5.68 + 2*9.16 = 24
5.68 * 9.16 = 52

2007-12-11 12:16:14 · answer #1 · answered by Anonymous · 1 0

The total width of the required sheet of aluminum is:

2D + W = 24

where D is the depth of the gutter and W is its width.

The cross sectional area of the gutter is D x W = 52

You have two equations in two unknowns so you can set up a quadratic and solve it.

2007-12-11 12:02:26 · answer #2 · answered by simplicitus 7 · 0 0

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