Question

A gutter is to be made out of a long strip of copper 9 in. wide by bending the strip along two lines parallel to the edges and distant respectively 3 in. from an edge. Thus the cross-section will be a broken line, made up of three straight lines, each 3 in. long. How wide should the gutter be at the top, in order that its carrying capacity may be as great as possible?

Short answer

Answer: The width of the gutter at the top should be approximately 2.61 inches to achieve maximum carrying capacity.

Step-by-step solution

Define the dimensions of the gutter and the trapezoid

Let the width of the gutter at the top be 2x and the height be h. Since both sides of the gutter are 3 inches long, the width at the bottom of the gutter is (9 - 2h) inches. The cross-section of the gutter is a trapezoid with height h and bases 2x and (9 - 2h).

Calculate the area of the trapezoid

To find the area of the trapezoid, we can use the following formula: Area = (1/2) x (sum of the bases) x height A = (1/2)(2x + 9 - 2h)h

Express the height h in terms of x using Pythagorean theorem

Using a right triangle that has the height h, side length 3, and half of the width at the top x as its sides, we can use the Pythagorean theorem to express h in terms of x: x^2 + h^2 = 3^2 Therefore, h^2 = 9 - x^2 h = sqrt(9-x^2)

Substitute the expression of h in terms of x back into the area formula

A = (1/2)(2x + 9 - 2(sqrt(9-x^2)))sqrt(9-x^2) Now, we will simplify the equation: A = x(4.5 - sqrt(9-x^2))

Find the maximum area using calculus

To find the maximum area, take the derivative of A with respect to x and set it to zero: dA/dx = (4.5 - sqrt(9-x^2)) - (x^2/(2sqrt(9-x^2))) Set dA/dx = 0: 0 = (4.5 - sqrt(9-x^2)) - (x^2/(2sqrt(9-x^2)))

Solve for x

To find the value of x that results in the maximum area, we solve the equation: (4.5 - sqrt(9-x^2))^2 = x^2/2 x ≈ 1.305 The width of the gutter at the top should be approximately 2x = 2(1.305) = 2.61 inches for its carrying capacity to be as great as possible.

Key concepts

Calculus Application

When solving optimization problems, calculus is an essential tool. It helps us to find the values that maximize or minimize a function. In the context of the gutter problem, our goal is to maximize the carrying capacity, which is directly related to the area of the trapezoidal cross-section.

To achieve this, we first express the area as a function of one variable, e.g., the width at the top (x). Calculus allows us to find critical points of this function by taking the derivative and setting it to zero. After finding the derivative, solving the equation gives us the x-value at which the area is greatest.

At its core, this process involves:

• Deriving a function to represent the problem.
• Finding the derivative to locate any potential maximums or minimums.
• Evaluating potential solutions in the context of your real-world problem.

Trapezoidal Area

The cross-section of the gutter forms a trapezoid, an important shape in geometry and calculus applications. The area of a trapezoid can be calculated using a specific formula which considers both of its bases and height.

In our problem, we define the trapezoid by the lengths 2x and (9 - 2h) as the bases, and h as the height. The formula for the area of a trapezoid is:

• Area = \( \frac{1}{2} \times (\text{sum of the bases}) \times \text{height} \)

Therefore, understanding how to properly substitute these values into the formula is crucial. This gives us an expression for the cross-section area, which we can then use to find the optimal size of the gutter.

This knowledge allows students to explore various geometric shapes and their properties related to real-world applications like optimizing space or material use.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry, which describes the relationship between the sides of a right triangle. This theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In the context of the gutter question, we use this theorem to express the height (h) of the trapezoid in terms of the top width (x). If we denote the side of the right triangle as h and half-width at the top as x, then the triangle's hypotenuse, which is the side of the bent gutter, will be 3 inches.

This gives us the formula:

• \( x^2 + h^2 = 3^2 \)
• Solving for h gives us \( h = \sqrt{9-x^2} \)

By substituting the expression for h back into the equation for the area, students can see how geometry and algebra are interwoven to solve real-life optimization problems.