### Problem StatementA box of volume $60 \, \text{m}^3$ with a square bottom and no top is constructed out of 2 different materials. The material for the bottom costs $ \$40/\text{m}^2 $. The material for the sides costs $ \$30/\text{m}^2 $.Find the dimensions that minimize total cost. (Don't forget to show that your answer is a minimum). Leave answer exact or give approximation to 2 decimal places.### Explanation- **Volume of the Box:** $ V = 60 \, \text{m}^3 $- **Cost per Unit Area for Bottom Material:** $ \$40/\text{m}^2 $- **Cost per Unit Area for Side Material:** $ \$30/\text{m}^2 $### Step-by-Step Solution1. **Determine the Variables:** - Let $ x $ be the length of one side of the square base. - Let $ h $ be the height of the box.2. **Establish the Volume Constraint:** - Volume: $ x^2 h = 60 \Rightarrow h = \frac{60}{x^2} $3. **Calculate the Surface Area:** - Bottom area: $ x^2 $ - Side area: $ 4xh $4. **Express Total Cost:** - Cost for bottom: $ 40x^2 $ - Cost for sides: $ 30 \cdot 4xh = 120xh $ - Substitute for $ h $: $ 120x \left( \frac{60}{x^2} \right) = \frac{7200}{x} $ - Total cost: $ C(x) = 40x^2 + \frac{7200}{x} $5. **Find the Minimum Cost:** - Derive the cost function with respect to $ x $: $ C'(x) = 80x - \frac{7200}{x^2} $ - Set the derivative to zero to find critical points: \( 80x - \frac{7200}{x^2} = 0 \Rightarrow 80x^3 = 7200 \Rightarrow x^3 = 90 \Rightarrow x = \sqrt

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Description: ### Problem StatementA box of volume $60 \, \text{m}^3$ with a square bottom and no top is constructed out of 2 different materials. The material for the bottom costs $ \$40/\text{m}^2 $. The material for the sides costs $ \$30/\text{m}^2 $.Fi

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Abox of volume 60m3 with a squan botHom and no top IS constructd out of 2 diffennt materi als. The makerial for the botHom costs $40/m². The makri al for the sides costs $30/m? And the dimensions that minimalize total cost. (don't forget to Show that answer is a minimum) Leare answer exact or qire approximahon to 2 decimal places.

Abox of volume 60m3 with a squan botHom and no top IS constructd out of 2 diffennt materi als. The makerial for the botHom costs $40/m². The makri al for the sides costs $30/m? And the dimensions that minimalize total cost. (don't forget to Show that answer is a minimum) Leare answer exact or qire approximahon to 2 decimal places.

Algebra for College Students

10th Edition

ISBN:9781285195780

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter9: Polynomial And Rational Functions

Section9.4: Graphing Polynomial Functions

Problem 43PS

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Question

### Problem Statement

A box of volume $60 \, \text{m}^3$ with a square bottom and no top is constructed out of 2 different materials. The material for the bottom costs $ \$40/\text{m}^2 $. The material for the sides costs $ \$30/\text{m}^2 $.

Find the dimensions that minimize total cost. (Don't forget to show that your answer is a minimum). Leave answer exact or give approximation to 2 decimal places.

### Explanation

- **Volume of the Box:** $ V = 60 \, \text{m}^3 $
- **Cost per Unit Area for Bottom Material:** $ \$40/\text{m}^2 $
- **Cost per Unit Area for Side Material:** $ \$30/\text{m}^2 $

### Step-by-Step Solution

1. **Determine the Variables:**
- Let $ x $ be the length of one side of the square base.
- Let $ h $ be the height of the box.

2. **Establish the Volume Constraint:**
- Volume: $ x^2 h = 60 \Rightarrow h = \frac{60}{x^2} $

3. **Calculate the Surface Area:**
- Bottom area: $ x^2 $
- Side area: $ 4xh $

4. **Express Total Cost:**
- Cost for bottom: $ 40x^2 $
- Cost for sides: $ 30 \cdot 4xh = 120xh $
- Substitute for $ h $: $ 120x \left( \frac{60}{x^2} \right) = \frac{7200}{x} $
- Total cost: $ C(x) = 40x^2 + \frac{7200}{x} $

5. **Find the Minimum Cost:**
- Derive the cost function with respect to $ x $: $ C'(x) = 80x - \frac{7200}{x^2} $
- Set the derivative to zero to find critical points: \( 80x - \frac{7200}{x^2} = 0 \Rightarrow 80x^3 = 7200 \Rightarrow x^3 = 90 \Rightarrow x = \sqrt

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