Question

A roof gutter is to be nade from a long strip of shect metal, \(24 \mathrm{~cm}\) wide, by bending up cqual amounts at each side through equal angles. Find the angle and the dintension that will make the carrying capacity of the sutter as large as nossible.

Short answer

The optimal angle is 45 degrees, and the width should be 12 cm.

Step-by-step solution

Define the Variables

Let the width of each bent-up side be x cm and the bending angle be θ.

Express the Cross-Section of the Gutter

The gutter forms an isosceles trapezoid with bases of length 24 cm - 2x.

Determine the Area of the Trapezoid

The cross-sectional area A of the trapezoid is given by the formula \( A = ((24 - 2x) \sin \theta) \cdot x \).

Find the Derivative

Differentiate the area function A with respect to x to find when the area is maximized.

Set the Derivative to Zero

Set the derivative equation to 0 and solve for x and \( \theta \) to find the maximum area.

Key concepts

Calculus Application

In calculus, optimization problems are crucial. They often involve finding the maximum or minimum value of a function. This type of problem is common in many practical situations, such as designing objects or optimizing resources.
In this exercise, we're dealing with maximizing the area of a cross-section of a gutter. This involves understanding shapes, applying calculus to find derivatives, and using critical points to determine maximum values.

• We need to identify which quantity to maximize - here, it's the area of the gutter.
• Use variables to describe the problem mathematically.
• Apply calculus techniques to find the optimum values.

Trapezoid Area Calculation

To find the area of the gutter, we need to understand the shape formed when the metal is bent. The gutter's cross-section is an isosceles trapezoid.

• Isosceles trapezoid: A trapezoid with equal angled sides.
• Area formula: For a trapezoid, the area is calculated by the formula \( A = ((a + b)/2) * h \), where \(a\text{ and }b \) are the lengths of the parallel sides, and \(h \) is the height.

In our exercise:

• Length of the bases: \(24 \text{ cm} - 2x \)
• Height: \(x \sin \theta \)
• Area formula used: \( A = ((24 - 2x) * x \sin \theta ) \)

Derivative for Maximization

To find the area where it's maximized, we need to use derivatives.

• First, we take the derivative of the area function \((A) \) with respect to the variable \((x)\).
• Set the derivative equal to zero to find critical points where the function's slope is zero. This can indicate maxima or minima.
• Solve the equation to find the values of \(x \text{ and } \theta\text{ for the maximum area.}\)

In this case, differentiating the area function helps identify the point where the area of the trapezoid is largest.

Mathematical Problem Solving

In mathematical problem-solving, we follow specific steps to achieve our goal.

• Define the problem clearly and identify the goal.
• Introduce variables to transform the problem into mathematical expressions.
• Apply the correct mathematical methods (like differentiation).
• Solve the resulting equations to find the values which give the desired outcome.

This roof gutter problem is a classical example where practical design meets mathematical theory. By expressing the gutter's area using trigonometric and algebraic expressions and applying calculus to find maximum area, we achieve an optimal solution to a real-world problem.