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Chapter 9: Problem 61

Dimensions of a Box A box with an open top has a square base and four sides of equal height. The volume of the box is \(360 \mathrm{ft}^{3} .\) If the surface area is \(276 \mathrm{ft}^{2},\) find the dimensions of the box. (Round answers to the nearest thousandth, if necessary.) (Figure can't copy)

Short Answer

Expert verified
The dimensions of the box are approximately 6 feet by 6 feet by 10 feet.

Step by step solution

01

Define variables

Let's denote the side length of the square base by \(s\) (in feet) and the height of the box by \(h\) (in feet).
02

Write expressions for volume and surface area

The volume of the box is given by: \[ V = s^2 \times h \]It is given that the volume is 360 ft\textsuperscript{3}. Therefore, \[ s^2 \times h = 360 \]
03

Express height in terms of side length

We can solve for height (\(h\)) in terms of side length (\(s\)) from the volume equation:\[ h = \frac{360}{s^2} \]
04

Write expression for surface area

The surface area of the box is given by:\[ \text{Surface Area} = s^2 + 4sh \]It is given that the surface area is 276 ft\textsuperscript{2}. Therefore,\[ s^2 + 4sh = 276 \]
05

Substitute height in the surface area equation

Substitute the expression for \(h\) from Step 3 into the surface area equation:\[ s^2 + 4s \left( \frac{360}{s^2} \right) = 276 \]This simplifies to:\[ s^2 + \frac{1440}{s} = 276 \]
06

Solve for side length (s)

Multiply through by \(s\) to clear the fraction:\[ s^3 + 1440 = 276s \]Rearrange to form a polynomial equation:\[ s^3 - 276s + 1440 = 0 \]This cubic equation can be solved using numerical methods or approximations.
07

Find the roots using approximation tools

Using a numerical solver or a graphing calculator, approximate the root of the cubic equation. The root near which to find the solution is approximately\(s \approx 6\).
08

Calculate height

Substitute the approximate value of \(s = 6\) into the expression for height:\[ h = \frac{360}{s^2} = \frac{360}{6^2} = \frac{360}{36} = 10 \]
09

Verify the dimensions

Double-check the calculations for both the volume and surface area to confirm that they match the given values: - Volume: \[ V = s^2 \times h = 6^2 \times 10 = 360 \] - Surface Area: \[ A = s^2 + 4sh = 6^2 + 4 \times 6 \times 10 = 36 + 240 = 276 \]

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Key Concepts

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

Volume of Box
The volume of a box is the amount of space it occupies. For a box with a square base and an open top, the volume can be calculated using the formula: \[ V = s^2 \times h \] where \(V\) represents the volume, \(s\) is the side length of the square base, and \(h\) is the height of the box. In this problem, we know that the volume is 360 cubic feet. This means: \[ s^2 \times h = 360 \] The volume formula helps us relate the dimensions of the box and ultimately find the height when the side length is known.
Surface Area
Surface area is the total area of all exposed surfaces of a three-dimensional object. For our box with one open top, the surface area formula is: \[ \text{Surface Area} = s^2 + 4sh \] Here, \(s^2\) is the area of the square base, and \(4sh\) represents the area of the four vertical sides. We know the surface area is 276 square feet, so we get: \[ s^2 + 4sh = 276 \] By substituting the expression for height from the volume calculation (\( h = \frac{360}{s^2} \)) into this equation, we can solve for the side length \(s\).
Cubic Equations
A cubic equation is a polynomial equation of the form \(ax^3 + bx^2 + cx + d = 0 \). In our exercise, an equation in this form arises when we simplify the combined surface area formula: \[ s^2 + \frac{1440}{s} = 276 \] By multiplying through by \(s\) to clear the fraction, we get: \[ s^3 - 276s + 1440 = 0 \] Solving cubic equations typically involves finding the roots that satisfy the equation. These can be solved using methods like factoring, using the Rational Root Theorem, or numerical approximations when exact solutions are difficult to obtain.
Approximation Methods
Exact solutions for cubic equations are often challenging to find, so approximation methods are commonly used. Some of these include:
  • Graphical Solutions: Plotting the equation and identifying where it crosses the x-axis.
  • Numerical Solvers: Using algorithms or calculators to estimate roots.
  • Trial and Error: Testing potential root values and refining them.
In this problem, a numerical solver or approximation shows that a root close to 6 meets the criteria: \[ s \approx 6 \] Substituting \(s = 6\) into the height equation: \[ h = \frac{360}{s^2} = \frac{360}{36} = 10 \] This gives us the dimensions, which match the given conditions when verified.

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