2. An open box is to be made from a square piece of material by cutting equal squares from each corner and turning up the sides. Find the dimensions of the box of maximum volume if the material has dimensions 6 in. by 6 in.

Mathematics For Machine Technology
8th Edition
ISBN:9781337798310
Author:Peterson, John.
Publisher:Peterson, John.
Chapter59: Areas Of Rectangles, Parallelograms, And Trapezoids
Section: Chapter Questions
Problem 79A
Question
**Problem 2:**

An open box is to be made from a square piece of material by cutting equal squares from each corner and turning up the sides. Find the dimensions of the box of maximum volume if the material has dimensions 6 inches by 6 inches.

*Explanation:*

- **Objective:** To determine the size of the squares to cut from each corner to maximize the volume of the resulting box.
- **Given:** A 6-inch by 6-inch square piece of material.
- **Process:** Calculate the volume of the box in terms of the size of the cut squares and maximize this volume.

**Steps for Solution:**

1. **Define Variables:** Let \( x \) be the side length of the square cut from each corner.
2. **Volume Equation:** Volume \( V \) of the box is given by: 
   \[
   V = x(6 - 2x)(6 - 2x)
   \]
3. **Maximize Volume:** Find \( x \) that maximizes \( V \).

**Study Tip:**

- Apply calculus techniques such as differentiation to find maximum points.
- Remember to consider the domain of \( x \), since \( x \) must be positive and \( 2x \leq 6 \).
Transcribed Image Text:**Problem 2:** An open box is to be made from a square piece of material by cutting equal squares from each corner and turning up the sides. Find the dimensions of the box of maximum volume if the material has dimensions 6 inches by 6 inches. *Explanation:* - **Objective:** To determine the size of the squares to cut from each corner to maximize the volume of the resulting box. - **Given:** A 6-inch by 6-inch square piece of material. - **Process:** Calculate the volume of the box in terms of the size of the cut squares and maximize this volume. **Steps for Solution:** 1. **Define Variables:** Let \( x \) be the side length of the square cut from each corner. 2. **Volume Equation:** Volume \( V \) of the box is given by: \[ V = x(6 - 2x)(6 - 2x) \] 3. **Maximize Volume:** Find \( x \) that maximizes \( V \). **Study Tip:** - Apply calculus techniques such as differentiation to find maximum points. - Remember to consider the domain of \( x \), since \( x \) must be positive and \( 2x \leq 6 \).
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