Problem 61 - Algebra 1: Concepts and Skills

Question

The length \(\ell\) of a box is 3 inches less than the height \(h .\) The width \(w\) is 9 inches less than the height. The box has a volume of 324 cubic inches. What are the dimensions of the box?

Step-by-Step Solution

Answer

The dimensions of the box are approximately: length \(\ell \approx 6\) inches, width \(w \approx 0\) inches, and height \(h \approx 9\) inches.

Step 1: Setup the equations from given conditions

We have three unknowns \(\ell\), \(w\), and \(h\). From the conditions given, we create two equations: \(\ell = h - 3\) and \(w = h - 9\). The third equation is from the volume of the box, \(V = \ell \cdot w \cdot h\) where \(V = 324\).

Step 2: Substitute the expressions of \(\ell\) and \(w\) into the volume Equation

Substitute \(\ell = h-3\) and \(w = h-9\) into the volume equation \(V = \ell \cdot w \cdot h\) we get \(V = (h - 3) \cdot (h - 9) \cdot h\). Substitute \(V = 324\) we get \(324 = (h - 3) \cdot (h - 9) \cdot h = h^3 - 12h^2 +27h\)

Step 3: Solve for \(h\) in the cubic equation

Setting the right side to zero, we have the equation \(h^3 - 12h^2 + 27h - 324 = 0\). Solving this equation using methods of factoring or calculus or graphing calculator, we find \(h \approx 9\).

Step 4: Substitute \(h \approx 9\) into equitations to find \(\ell\) and \(w\)

Substitute \(h \approx 9\) into equations \(\ell = h - 3\) and \(w = h - 9\) to solve for \(\ell\) and \(w\). After substitution we find \(\ell \approx 6\) inches and \(w \approx 0\) inches.

Key Concepts

Solving Cubic Equations

Understanding how to solve cubic equations is essential for tackling a range of mathematical problems, including finding the dimensions of three-dimensional shapes such as a box. A cubic equation is a polynomial equation of the third degree; the general form is \( ax^3 + bx^2 + cx + d = 0 \). The methods of solving these include factoring by grouping, using the rational root theorem, or approximation methods such as Newton's method if the roots are difficult to find analytically.

For the problem at hand, we focused on the cubic equation \( h^3 - 12h^2 + 27h - 324 = 0 \) derived from the volume condition. To factor such an equation, you want to look for patterns or make an educated guess about possible roots. If a cubic equation doesn't factor easily, graphing calculators or numerical methods can be great tools to estimate the solution. Step 3 of the solution finds the height to be approximately 9 inches, which is crucial for calculating the box's dimensions.

Systems of Equations

A system of equations is a set of multiple equations, usually involving the same variables, which we aim to solve together. They can be linear or non-linear, and in this problem, we deal with a linear system when setting the relationships between the dimensions of the box. To fully determine the box's dimensions, we are given that the length \( \ell \) is 3 inches less than the height \( h \) and the width \( w \) is 9 inches less than the height. This gives us two equations: \( \ell = h - 3 \) and \( w = h - 9 \).

Such systems can typically be solved by substitution or elimination methods. As we already have \( \ell \) and \( w \) expressed in terms of \( h \), the substitution into the volume equation is straightforward, leading to the cubic equation we previously discussed. The key is to express all variables in terms of a single variable to solve the system efficiently.

Volume Calculation

The volume of a box, which is a three-dimensional object, is found by multiplying the length, width, and height. The formula is: \( V = \ell \cdot w \cdot h \), where \( V \) is the volume, \( \ell \) is the length, \( w \) is the width, and \( h \) is the height of the box. In mathematical problems, the volume might be provided, and the task is to work backward to find the dimensions.

It's crucial to notice that a volume calculation is a particular case of solving a system of equations — one where we know the product of the variables involved. In the problem given, we aimed to find the dimensions that conform to the volume of 324 cubic inches. This condition turned out to be the bridge between an abstract cubic equation and a practical problem, highlighting the importance of understanding volume calculations in algebra.