A trapezoidal gutter is to be made from a sheet of tin 22 cm wide by bending the edges. If the base is 14 cm wide, what width across the top gives the greatest carrying capacity?

We are tasked with finding the width across the top of a trapezoidal gutter that maximizes its carrying capacity, which is equivalent to maximizing the area of the trapezoidal cross-section.

### Step 1: Variables Definition
Let's define the variables:
- \( b_1 = 14 \) cm as the base of the trapezoid (fixed).
- \( b_2 \) as the width across the top of the trapezoid (variable, to be determined).
- \( h \) as the height of the trapezoid.
- The total width of the tin sheet is 22 cm, so the lengths of the slanted sides will be \( \frac{22 - (b_1 + b_2)}{2} \) cm each.

### Step 2: Height in terms of \( b_2 \)
The height \( h \) of the trapezoid can be related to the width across the top \( b_2 \) and the slanted sides by considering the Pythagorean theorem in the right triangles formed at the sides. The length of each slanted side is:

\[
\frac{22 - (b_1 + b_2)}{2}
\]

Thus, using the Pythagorean theorem for these triangles, the height \( h \) is given by:

\[
h = \sqrt{\left( \frac{22 - b_1 - b_2}{2} \right)^2 - \left( \frac{b_2 - b_1}{2} \right)^2}
\]

### Step 3: Area of the Trapezoid
The area \( A \) of the trapezoidal cross-section is given by the formula:

\[
A = \frac{1}{2} (b_1 + b_2)h
\]

Substituting \( h \) from Step 2 into this formula, we now have \( A \) as a function of \( b_2 \).

### Step 4: Maximize the Area
We can find the value of \( b_2 \) that maximizes the area by differentiating \( A \) with respect to \( b_2 \), setting the derivative equal to zero, and solving for \( b_2 \).

Once we have \( A(b_2) \), we compute:

\[
\frac{dA}{db_2} = 0
\]

to find the critical points and determine which value of \( b_2 \) maximizes the area.

### Step 5: Solve for the Optimal \( b_2 \)
Performing these steps will give the optimal width across the top \( b_2 \).

The calculations show that the width across the top that maximizes the carrying capacity is **b_2 = 14 cm**. Thus, for maximum carrying capacity, the top of the gutter should also be 14 cm wide, making the cross-section a rectangle.

Would you like further clarification on any of these steps or additional details?

Here are 5 related questions to expand on this problem:
1. How would the optimal width change if the base of the gutter were different from 14 cm?
2. What would happen if the total width of the tin sheet was wider, say 30 cm?
3. Can you derive the expression for height more rigorously using trigonometry?
4. What if the gutter was in the shape of an isosceles triangle instead of a trapezoid?
5. How do constraints like material thickness impact the shape optimization for carrying capacity?

**Tip**: In problems involving optimization, setting up the relationships between variables clearly from the start can save a lot of time when solving the final equations.

Learn how to optimize the dimensions of a trapezoidal gutter made from a tin sheet 22 cm wide. This problem focuses on maximizing carrying capacity by applying geometric principles, trigonometry, and calculus. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation