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A rain gutter is made from sheets of aluminum that are 27 inches wide. The edges are turned up to form right angles. Determine the depth of the gutter that will allow a cross-sectional area of 54 square inches. There are two solutions to this problem. Round to the nearest tenth of an inch.




a. 1.7 in. and 19.9 in.

b. 2.2 in. and 24.8 in.

c. 2.9 in. and 13.3 in.

d. 2.4 in. and 11.1 in.

Sagot :

Representing a situation as an algebraic equation using variables can be used to solve complex problems

The correct option as option for the depth of the gutter that will allow a cross-sectional area of 54 square inches as option d.

d. y ≈ 2.4 in. and y ≈ 11.1 in.

The reason the selected option is correct is as follows:

The given parameters of the rain gutter are;

The width of the aluminum sheets from which the rain gutter is made = 27 inches

The angle to which the edges are turned up to form right angles = Right angles (90 degrees)

Required:

To determine the depth of the gutter allowing a cross-sectional area of 54 square inches

Method:

Form an equation that describe the area of the gutter, taking note that the gutter does not have a cover (top) and therefore only three sides are required

Solution:

Let y represent the length of the sides (depth) of the rain gutter, and let x represent the width, we have;

2·y + x = 27...(1) (The total with of the gutter; 2 sides and a base)

The cross sectional area of the gutter, A = y × x...(2)

Where;

y = The depth of the gutter

x = The width of the gutter

From equation (1), we have;

x = 27 - 2·y

Therefore, in equation (2), we have;

A = y × (27 - 2·y)

The desired cross-sectional area, A = 54 in.²

∴ A = 54 = y × (27 - 2·y) = 27·y - 2·y²

Which gives;

2·y² - 27·y + 54 = 0

From which we get;

[tex]y = \dfrac{27 \pm \sqrt{(-27)^2 - 4 \times 2 \times 54} }{2 \times 2} = \dfrac{27 \pm 3\times \sqrt{33} }{4}[/tex]

Therefore;

y ≈ 2.4 and/or y ≈ 11.1

Which gives the depth of the gutter that will allow a cross-sectional area of 54 square inches as y ≈ 2.4 in. and y ≈ 11.1 in.

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