A box with a square base and open top must have a volume of 157216 cm³. We wish to find the dimensions of the box that minimize the amount of material used. First, find a formula for the surface area of the box in terms of only z, the length of one side of the square base. [Hint: use the volume formula to express the height of the box in terms of x.] Simplify your formula as much as possible. A(x) = Next, find the derivative, A'(x). A'(x) = Now, calculate when the derivative equals zero, that is, when A'(x) = 0. [Hint: multiply both sides by 2².] A'(x) = 0 when x = 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)= Evaluate A"(x) at the x-value you gave above. = NOTE: Since your last answer is positive, this means that the graph of A(z) is concave up around that value, so the zero of A'(x) must indicate a local minimum for A(z). (Your boss is happy now.) Jump to Answer
A box with a square base and open top must have a volume of 157216 cm³. We wish to find the dimensions of the box that minimize the amount of material used. First, find a formula for the surface area of the box in terms of only z, the length of one side of the square base. [Hint: use the volume formula to express the height of the box in terms of x.] Simplify your formula as much as possible. A(x) = Next, find the derivative, A'(x). A'(x) = Now, calculate when the derivative equals zero, that is, when A'(x) = 0. [Hint: multiply both sides by 2².] A'(x) = 0 when x = 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)= Evaluate A"(x) at the x-value you gave above. = NOTE: Since your last answer is positive, this means that the graph of A(z) is concave up around that value, so the zero of A'(x) must indicate a local minimum for A(z). (Your boss is happy now.) Jump to Answer
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![The image contains a mathematical problem about optimizing the dimensions of a box with a square base and an open top to minimize the amount of material used. The box is specified to have a volume of 157216 cm³.
### Problem Statement:
- **Objective:** Minimize the surface area of the box.
- **Given:** Volume of the box = 157216 cm³.
### Steps to Solve:
1. **Formula for Surface Area (A(x)):**
- Derive the formula for the surface area of the box in terms of \( x \), where \( x \) is the length of one side of the square base.
- Use the volume formula to express the height in terms of \( x \).
- Simplify the expression for \( A(x) \).
\[
A(x) =
\]
2. **Find the Derivative (A′(x)):**
- Calculate the derivative of \( A(x) \).
\[
A′(x) =
\]
3. **Set Derivative to Zero:**
- Solve for when the derivative equals zero to find potential critical points. Multiply both sides by \( x^2 \).
\[
A′(x) = 0 \text{ when } x =
\]
4. **Ensure Minimum Value:**
- Use the second derivative test to confirm that the value of \( x \) found provides a minimum surface area.
- Calculate the second derivative, \( A″(x) \).
\[
A″(x) =
\]
5. **Evaluate the Second Derivative:**
- Assess \( A″(x) \) at the previously calculated \( x \)-value.
### Note:
- If the second derivative is positive, the function \( A(x) \) is concave up at that value, indicating a local minimum for the surface area.
- The diagram or formula in this context is the series of calculations and logical steps required to find the optimal dimensions for the box.
### Submission Buttons:
- **Submit Question**
- **Jump to Answer**
### Additional Notes:
- Ensure your inputs are consistent and logical throughout the problem-solving process to achieve accurate results.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F91f806a5-3485-4560-baf7-fe0d7b379a18%2F856663b1-c636-42c2-b86c-cf534872b8cf%2F1i71k0f7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image contains a mathematical problem about optimizing the dimensions of a box with a square base and an open top to minimize the amount of material used. The box is specified to have a volume of 157216 cm³.
### Problem Statement:
- **Objective:** Minimize the surface area of the box.
- **Given:** Volume of the box = 157216 cm³.
### Steps to Solve:
1. **Formula for Surface Area (A(x)):**
- Derive the formula for the surface area of the box in terms of \( x \), where \( x \) is the length of one side of the square base.
- Use the volume formula to express the height in terms of \( x \).
- Simplify the expression for \( A(x) \).
\[
A(x) =
\]
2. **Find the Derivative (A′(x)):**
- Calculate the derivative of \( A(x) \).
\[
A′(x) =
\]
3. **Set Derivative to Zero:**
- Solve for when the derivative equals zero to find potential critical points. Multiply both sides by \( x^2 \).
\[
A′(x) = 0 \text{ when } x =
\]
4. **Ensure Minimum Value:**
- Use the second derivative test to confirm that the value of \( x \) found provides a minimum surface area.
- Calculate the second derivative, \( A″(x) \).
\[
A″(x) =
\]
5. **Evaluate the Second Derivative:**
- Assess \( A″(x) \) at the previously calculated \( x \)-value.
### Note:
- If the second derivative is positive, the function \( A(x) \) is concave up at that value, indicating a local minimum for the surface area.
- The diagram or formula in this context is the series of calculations and logical steps required to find the optimal dimensions for the box.
### Submission Buttons:
- **Submit Question**
- **Jump to Answer**
### Additional Notes:
- Ensure your inputs are consistent and logical throughout the problem-solving process to achieve accurate results.
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