A box with a square base and open top must have a volume of 296352 cm³. A(x) We wish to find the dimensions of the box that minimize the amount of material used. Step 1: Find a formula for the surface area of the box in terms of only x, the length of one side of the square base. Hint: use the volume formula to express the height, y of the box in terms of x, and then substitute this into the surface area formula. Simplify your formula as much as possible. = x //////// Step 2: Find the derivative, A'(x). A'(x) = /////// x Find the critical values, that is solve A'(x) = 0 for x. Hint: Multiply both sides for ² The critical value(s) are x = Y We next have to make sure that this value of a gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x). A"(x) = and then x = y = Evaluate A"(x) at the x-value you gave above. The dimensions of the box that minimizes the volume is Part 3 of 3
A box with a square base and open top must have a volume of 296352 cm³. A(x) We wish to find the dimensions of the box that minimize the amount of material used. Step 1: Find a formula for the surface area of the box in terms of only x, the length of one side of the square base. Hint: use the volume formula to express the height, y of the box in terms of x, and then substitute this into the surface area formula. Simplify your formula as much as possible. = x //////// Step 2: Find the derivative, A'(x). A'(x) = /////// x Find the critical values, that is solve A'(x) = 0 for x. Hint: Multiply both sides for ² The critical value(s) are x = Y We next have to make sure that this value of a gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x). A"(x) = and then x = y = Evaluate A"(x) at the x-value you gave above. The dimensions of the box that minimizes the volume is Part 3 of 3
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![**Title: Minimizing Material for a Box with an Open Top**
**Objective:**
Determine the dimensions of a box with a square base and open top to minimize the material used while maintaining a volume of \(296352 \, \text{cm}^3\).
---
### Diagram:
- A 3D diagram of a box with a square base is shown. The sides of the square base are labeled \(x\), and the height of the box is labeled \(y\).
### Process:
#### Step 1: Express the Surface Area
- **Goal:** Formulate the surface area \(A(x)\) in terms of the side length \(x\).
- **Hint:** Use the box's volume to express the height \(y\) in terms of \(x\), and substitute into the surface area formula.
- **Volume Formula:** \(x^2y = 296352\)
- **Simplify:** \(A(x) = \) [Input Formula]
#### Step 2: Find the Derivative
- **Objective:** Calculate the derivative \(A'(x)\).
- **Derivative:** \(A'(x) = \) [Input Derivative]
#### Step 3: Determine Critical Values
- **Task:** Solve \(A'(x) = 0\) to find critical values for \(x\).
- **Hint:** Multiply both sides for \(x^2\).
- **Critical Value:** \(x = \) [Input Critical Value]
#### Step 4: Confirm Minimum Surface Area
- **Approach:** Use the second derivative test to confirm minimum surface area.
- **Second Derivative:** \(A''(x) = \) [Input Second Derivative]
- **Evaluate:** Compute \(A''(x)\) at the critical value.
### Conclusion
- **Dimensions that Minimize Material:**
- \(x =\) [Input x Value]
- \(y =\) [Input y Value]
**Reminder:** The steps above guide you through finding the optimal dimensions for minimizing material usage while maintaining a given volume. Use calculus principles for differentiation and applying tests for minima to achieve the solution.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F05c42395-925d-4e90-8ff6-49150fb199bf%2F347e30cc-ce9f-487a-a236-e8ea3ba72044%2Fypbjxb_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Minimizing Material for a Box with an Open Top**
**Objective:**
Determine the dimensions of a box with a square base and open top to minimize the material used while maintaining a volume of \(296352 \, \text{cm}^3\).
---
### Diagram:
- A 3D diagram of a box with a square base is shown. The sides of the square base are labeled \(x\), and the height of the box is labeled \(y\).
### Process:
#### Step 1: Express the Surface Area
- **Goal:** Formulate the surface area \(A(x)\) in terms of the side length \(x\).
- **Hint:** Use the box's volume to express the height \(y\) in terms of \(x\), and substitute into the surface area formula.
- **Volume Formula:** \(x^2y = 296352\)
- **Simplify:** \(A(x) = \) [Input Formula]
#### Step 2: Find the Derivative
- **Objective:** Calculate the derivative \(A'(x)\).
- **Derivative:** \(A'(x) = \) [Input Derivative]
#### Step 3: Determine Critical Values
- **Task:** Solve \(A'(x) = 0\) to find critical values for \(x\).
- **Hint:** Multiply both sides for \(x^2\).
- **Critical Value:** \(x = \) [Input Critical Value]
#### Step 4: Confirm Minimum Surface Area
- **Approach:** Use the second derivative test to confirm minimum surface area.
- **Second Derivative:** \(A''(x) = \) [Input Second Derivative]
- **Evaluate:** Compute \(A''(x)\) at the critical value.
### Conclusion
- **Dimensions that Minimize Material:**
- \(x =\) [Input x Value]
- \(y =\) [Input y Value]
**Reminder:** The steps above guide you through finding the optimal dimensions for minimizing material usage while maintaining a given volume. Use calculus principles for differentiation and applying tests for minima to achieve the solution.
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![### Minimizing the Material for an Open-Top Box
A box with a square base and an open top must have a volume of \(296352 \, \text{cm}^3\).
#### Objective
Determine the dimensions of the box that minimize the amount of material used.
#### Steps
**Step 1: Formulate the Surface Area**
Find a formula for the surface area of the box in terms of only \(x\), the length of one side of the square base.
- **Hint**: Use the volume formula to express the height, \(y\), of the box in terms of \(x\), and then substitute this into the surface area formula.
**Expression to Simplify:**
\[ A(x) = \]
**Step 2: Derivative Calculation**
Find the derivative, \( A'(x) \):
\[ A'(x) = \]
**Critical Value Calculation**
Identify the critical values by solving \( A'(x) = 0 \) for \( x \).
- **Hint**: Multiply both sides by \(x^2\).
The critical value(s) are:
\[ x = \]
#### Second Derivative Test
Ensure the critical value of \(x\) gives a minimum surface area by using the second derivative test.
\[ A''(x) = \]
Then, evaluate \( A''(x) \) at the \(x\)-value determined above.
---
By following these steps, you can find the optimal dimensions \(x\) and \(y\) for minimizing the material used in constructing the box while maintaining the required volume.](https://content.bartleby.com/qna-images/question/05c42395-925d-4e90-8ff6-49150fb199bf/4b6f35c8-c240-44cf-a1ea-1d1bed54ad60/lglbfs_processed.png)
Transcribed Image Text:### Minimizing the Material for an Open-Top Box
A box with a square base and an open top must have a volume of \(296352 \, \text{cm}^3\).
#### Objective
Determine the dimensions of the box that minimize the amount of material used.
#### Steps
**Step 1: Formulate the Surface Area**
Find a formula for the surface area of the box in terms of only \(x\), the length of one side of the square base.
- **Hint**: Use the volume formula to express the height, \(y\), of the box in terms of \(x\), and then substitute this into the surface area formula.
**Expression to Simplify:**
\[ A(x) = \]
**Step 2: Derivative Calculation**
Find the derivative, \( A'(x) \):
\[ A'(x) = \]
**Critical Value Calculation**
Identify the critical values by solving \( A'(x) = 0 \) for \( x \).
- **Hint**: Multiply both sides by \(x^2\).
The critical value(s) are:
\[ x = \]
#### Second Derivative Test
Ensure the critical value of \(x\) gives a minimum surface area by using the second derivative test.
\[ A''(x) = \]
Then, evaluate \( A''(x) \) at the \(x\)-value determined above.
---
By following these steps, you can find the optimal dimensions \(x\) and \(y\) for minimizing the material used in constructing the box while maintaining the required volume.
Solution
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