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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![### Optimization Problem: Maximum Volume of a Box
**Problem Statement:**
If 1,200 cm² of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
**Solution:**
To solve this problem, we will use concepts from calculus, specifically optimization. We will consider the dimensions of the box to optimize its volume given the surface area constraint.
1. Let the side of the square base be \( x \) cm and the height of the box be \( h \) cm.
2. The surface area constraint can be expressed as:
\[
x^2 + 4xh = 1200
\]
3. The volume \( V \) of the box is given by:
\[
V = x^2h
\]
4. Use the surface area equation to express \( h \) in terms of \( x \):
\[
h = \frac{1200 - x^2}{4x}
\]
5. Substitute the expression for \( h \) into the volume equation:
\[
V = x^2 \left( \frac{1200 - x^2}{4x} \right) = \frac{x(1200 - x^2)}{4}
\]
Simplify to:
\[
V = 300x - \frac{x^3}{4}
\]
6. To find the maximum volume, take the derivative \( V' \) and set it to zero:
\[
V' = 300 - \frac{3x^2}{4} = 0
\]
7. Solve for \( x \):
\[
\frac{3x^2}{4} = 300 \quad \Rightarrow \quad 3x^2 = 1200 \quad \Rightarrow \quad x^2 = 400 \quad \Rightarrow \quad x = 20
\]
8. Substitute \( x = 20 \) back into the equation for \( h \):
\[
h = \frac{1200 - 20^2}{4 \times 20} = \frac{1200 - 400}{80} = 10
\]
9. Calculate the volume:
\[
V = 20^2 \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61691e30-4fa2-4a1d-9914-fd132b93b22e%2F8dff6101-9aee-44fc-be4d-c42165d04c1b%2F1jr61l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Optimization Problem: Maximum Volume of a Box
**Problem Statement:**
If 1,200 cm² of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
**Solution:**
To solve this problem, we will use concepts from calculus, specifically optimization. We will consider the dimensions of the box to optimize its volume given the surface area constraint.
1. Let the side of the square base be \( x \) cm and the height of the box be \( h \) cm.
2. The surface area constraint can be expressed as:
\[
x^2 + 4xh = 1200
\]
3. The volume \( V \) of the box is given by:
\[
V = x^2h
\]
4. Use the surface area equation to express \( h \) in terms of \( x \):
\[
h = \frac{1200 - x^2}{4x}
\]
5. Substitute the expression for \( h \) into the volume equation:
\[
V = x^2 \left( \frac{1200 - x^2}{4x} \right) = \frac{x(1200 - x^2)}{4}
\]
Simplify to:
\[
V = 300x - \frac{x^3}{4}
\]
6. To find the maximum volume, take the derivative \( V' \) and set it to zero:
\[
V' = 300 - \frac{3x^2}{4} = 0
\]
7. Solve for \( x \):
\[
\frac{3x^2}{4} = 300 \quad \Rightarrow \quad 3x^2 = 1200 \quad \Rightarrow \quad x^2 = 400 \quad \Rightarrow \quad x = 20
\]
8. Substitute \( x = 20 \) back into the equation for \( h \):
\[
h = \frac{1200 - 20^2}{4 \times 20} = \frac{1200 - 400}{80} = 10
\]
9. Calculate the volume:
\[
V = 20^2 \times
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