C'(x) = Solving C'=0 yields that value of x that minimizes the cost. X=____inchesy=_____inches ### Problem Description:A box, with a square base, is to have a volume of $ V = 360 $ cubic inches. The cost for the materials of the four sides is $6 per square inch, while the cost of the material for the top and bottom is $10 per square inch.### Diagram Description:The diagram provided is a 3D representation of a rectangular box with one dimension labeled as $ x $ (the length of the side of the square base) and another dimension labeled as $ y $ (the height of the box). The other dimension of the square base is also indicated as $ x $.### Goals:1. Determine the dimensions $ x $ and $ y $ that minimize the total cost of the materials used to build the box.2. Calculate the total minimum cost based on these dimensions.### Formulas and Calculations:#### Volume of the Box:\[ V = x^2 \cdot y \]Given $ V = 360 $ cubic inches:\[ x^2 \cdot y = 360 \]\[ y = \frac{360}{x^2} \]#### Surface Area Calculations:- **Area of the Four Sides:** - Each side of the box has an area of $ x \cdot y $. - There are four sides, hence the total area for the sides is $ 4 \cdot (x \cdot y) = 4xy $.- **Area of the Top and Bottom:** - Each of the top and bottom has an area of $ x \cdot x = x^2 $. - There are two such surfaces, hence the total area for the top and bottom is $ 2x^2 $.#### Cost Calculations:- **Cost for the Four Sides:** \[ \text{Cost of sides} = 4xy \cdot 6 \]- **Cost for the Top and Bottom:** \[ \text{Cost of top and bottom} = 2x^2 \cdot 10 \]- **Total Cost (C):** \[ \text{Total cost} = 24xy + 20x^2 \]### Summary:To minimize the total cost of the materials, we need to determine the optimal values of $ x $ and $ y $ under the constraint \( x^

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A box, with a square base is to have a volume of V = 360 cubic inches. The cost for the materials of the four sides is $6 per square inch, while the cost of the material for the top and bottom is $10 per square inch. y X X

A box, with a square base is to have a volume of V = 360 cubic inches. The cost for the materials of the four sides is $6 per square inch, while the cost of the material for the top and bottom is $10 per square inch. y X X

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

100%

C'(x) =

Solving C'=0 yields that value of x that minimizes the cost. X=____inches

y=_____inches

### Problem Description:

A box, with a square base, is to have a volume of $ V = 360 $ cubic inches. The cost for the materials of the four sides is $6 per square inch, while the cost of the material for the top and bottom is $10 per square inch.

### Diagram Description:

The diagram provided is a 3D representation of a rectangular box with one dimension labeled as $ x $ (the length of the side of the square base) and another dimension labeled as $ y $ (the height of the box). The other dimension of the square base is also indicated as $ x $.

### Goals:

1. Determine the dimensions $ x $ and $ y $ that minimize the total cost of the materials used to build the box.
2. Calculate the total minimum cost based on these dimensions.

### Formulas and Calculations:

#### Volume of the Box:
\[ V = x^2 \cdot y \]
Given $ V = 360 $ cubic inches:
\[ x^2 \cdot y = 360 \]
\[ y = \frac{360}{x^2} \]

#### Surface Area Calculations:

- **Area of the Four Sides:**
- Each side of the box has an area of $ x \cdot y $.
- There are four sides, hence the total area for the sides is $ 4 \cdot (x \cdot y) = 4xy $.

- **Area of the Top and Bottom:**
- Each of the top and bottom has an area of $ x \cdot x = x^2 $.
- There are two such surfaces, hence the total area for the top and bottom is $ 2x^2 $.

#### Cost Calculations:

- **Cost for the Four Sides:**
\[ \text{Cost of sides} = 4xy \cdot 6 \]

- **Cost for the Top and Bottom:**
\[ \text{Cost of top and bottom} = 2x^2 \cdot 10 \]

- **Total Cost (C):**
\[ \text{Total cost} = 24xy + 20x^2 \]

### Summary:

To minimize the total cost of the materials, we need to determine the optimal values of $ x $ and $ y $ under the constraint \( x^

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