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To solve this problem of finding the dimensions of a rectangular box (without a top) with maximum capacity, given the total surface area of \( 108 \, \text{m}^2 \), we need to follow a standard optimization approach in calculus. Let's break it down:

### Problem Setup

1. Let the length of the base be \( l \), the width be \( w \), and the height be \( h \) of the box.
2. The surface area of the box consists of the base and four sides (since there's no top). The surface area formula for a box without a top is:
   \[
   A = l \cdot w + 2 \cdot l \cdot h + 2 \cdot w \cdot h
   \]
   We are given that the total surface area \( A \) is 108 m²:
   \[
   l \cdot w + 2 \cdot l \cdot h + 2 \cdot w \cdot h = 108
   \]

3. The volume \( V \) of the box is:
   \[
   V = l \cdot w \cdot h
   \]
   Our goal is to maximize the volume \( V \), subject to the surface area constraint.

### Steps to Solve:
1. Express \( h \) in terms of \( l \) and \( w \) using the surface area equation.
2. Substitute \( h \) into the volume equation.
3. Maximize the volume function by taking partial derivatives and setting them to zero (this will give the critical points).
4. Solve for the dimensions \( l \), \( w \), and \( h \) that give the maximum volume.

Let me calculate the exact dimensions by solving this system for you.The optimal dimensions of the rectangular box for maximum capacity, given a surface area of 108 m², are:

- Length \( l = 6 \, \text{m} \)
- Width \( w = 6 \, \text{m} \)
- Height \( h = 3 \, \text{m} \)

This configuration gives the maximum volume while satisfying the surface area constraint.

Would you like further details or have any questions? Here are five related questions to explore further:

1. How is the maximum volume of the box calculated based on these dimensions?
2. How would the dimensions change if the surface area were different?
3. What if the box had a top? How would that affect the optimization problem?
4. Can this method be applied to boxes of different shapes, like a cylinder or a triangular prism?
5. What are other real-world applications of optimization in surface area and volume problems?

**Tip:** When solving optimization problems with constraints, always start by expressing one variable in terms of others using the constraint equations. This reduces the complexity of the problem.

Obtain the dimensions of a rectangular box without top of maximum capacity given that the total surface area is 108m².

Solve for the dimensions of a rectangular box without a top that maximizes volume, given a total surface area of 108m². This optimization problem applies calculus concepts and explores surface area and volume relations for a box. Suitable for Grades 11-12.

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