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Chapter 4: Problem 68

A box with a square base and open top is to be made from sheet metal so that its volume is \(500 \mathrm{in}^{3}\). Let \(S\) be the area of the surface of the box and \(x\) the length of a side of the square base. Show that \(S=x^{2}+2000 / x\) for \(x>0\), and sketch the graph of \(S\) versus \(x\) for \(x>0\)

Short Answer

Expert verified
The surface area is given by \( S = x^2 + \frac{2000}{x} \), for \( x > 0 \).

Step by step solution

01

Understanding the Problem

We need to find the expression for the surface area \( S \) of a box with a square base and an open top. The box's volume is given as \( 500 \text{ in}^3 \), and \( x \) represents the length of one side of the base. We need to express \( S \) as a function of \( x \) and sketch the graph of \( S \) versus \( x \).
02

Set up the Volume Equation

The volume of the box is given by \( V = x^2h \), where \( h \) is the height of the box. We know that the volume is \( 500 \text{ in}^3 \). So, we have the equation \( x^2h = 500 \).
03

Solve for Height in Terms of \( x \)

From the volume equation \( x^2h = 500 \), we can solve for \( h \): \( h = \frac{500}{x^2} \). This will be used to express the surface area in terms of \( x \).
04

Derive the Surface Area Equation

The surface area \( S \) of the box consists of the area of the base and the four sides. The base has area \( x^2 \), and each side has area \( xh = x\frac{500}{x^2} = \frac{500}{x} \). There are four sides, so the total area of the sides is \( 4 \times \frac{500}{x} \). Thus, \( S = x^2 + 4 \times \frac{500}{x} = x^2 + \frac{2000}{x} \).
05

Confirm the Derived Expression

We have found that the surface area \( S = x^2 + \frac{2000}{x} \). This matches the problem statement requirement. This formula applies for \( x > 0 \).
06

Sketch the Graph of \( S \) Versus \( x \)

To sketch the graph of \( S \) versus \( x \), notice the behavior of the function. As \( x \) increases, \( x^2 \) increases while \( \frac{2000}{x} \) decreases, making \( S \) eventually increase. For very small \( x \), \( \frac{2000}{x} \) dominates and \( S \) becomes very large. For large \( x \), \( x^2 \) dominates. Plotting would show \( S \) dropping steeply from a high value, reaching a minimum, and then increasing again.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Area of a Box
Imagine you're designing a box with a square base and no lid. The task is to find out its surface area when the base length is known, and the box must hold a certain volume. This means:
  • Our box has a base area of \(x^2\) because both sides of the square base are of equal length \(x\).
  • Each of the four side panels has an area of \(xh\), where \(h\) is the height of the box.
To find the total surface area, you sum the area of the base with the areas of the four side panels. If the side panels are \(th\), the equation gets simplified with our volume constraint: \(S = x^2 + 4 \left(\frac{500}{x}\right)\). This formula means:
  • When \(x\) is small, the term \(\frac{2000}{x}\) makes a significant contribution.
  • When \(x\) is large, \(x^2\) dominates the surface area value.
Understanding this behavior helps in visualizing how the surface area changes as \(x\) varies.
Volume Constraint
Understanding constraints like volume is crucial in optimization problems. Here, the volume constraint ensures that the box can hold a specified volume, i.e., 500 cubic inches. This constrains the dimensions of the box.
Given a box with volume \(V = x^2h\) and knowing \(V = 500\), we can visualize the relationship between the base length \(x\) and height \(h\).
  • From, \(x^2h = 500\), we rearrange to find \(h = \frac{500}{x^2}\).
  • This tells us as \(x\) increases, \(h\) correspondingly decreases to maintain the volume constant.
This aspect of 'balancing' ensures the overall box shape varies but maintains its set volume. Calculating this also leads us to the surface area equation because knowing the volume enables expressing height as a function of \(x\).
Graph Sketching
Graph sketching is a valuable tool to visualize how the surface area changes with different base lengths. The function \(S = x^2 + \frac{2000}{x}\) sketches a curve that tells this story:
  • When \(x\) is small, \(\frac{2000}{x}\) is dominant, thus \(S\) is very high.
  • As \(x\) grows, \(x^2\) starts to take over, causing \(S\) first to drop.
  • Eventually, as \(x\) continues to increase, \(S\) rises again due to \(x^2\) getting larger.
The graph starts at a large surface area for small base lengths, dips to a minimum, and then ascends for larger bases. A plotted curve demonstrating these ups and downs can guide through analyzing optimization in real-world scenarios. It teaches us the trade-offs involved and predicts how slight adjustments can change total surface area while keeping the volume steady.

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