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?

Short answer

The dimensions of the box are length \(l = h - 3\), width \(w = h - 9\) and the height \(h\) which is found by solving the equation \(324 = (h-3) *(h - 9) * h\).

Step-by-step solution

Express all dimensions in terms of height

From the problem's description we know that: \(l = h - 3\) and \(w = h - 9\).

Set up the Volume equation

We are given that Volume of the box is 324 cubic inches. The volume for a box is given by \(Volume = l * w * h\). Substituting the expressions for l and w from step 1 into this equation, we get \(324 = (h-3) * (h-9) * h\).

Solve for 'h'

Solving the equation \(324 = (h-3) *(h - 9) * h\) will give us the value of height.

Solve for length 'l' and width 'w'

After finding 'h', substitute the value of 'h' into the equations \(l = h - 3\) and \(w = h - 9\) to get the values of length 'l' and width 'w'

Key concepts

Algebraic Expressions

Understanding algebraic expressions is a foundational aspect of solving many mathematical problems, especially those related to geometry and measurement. Let's dive into the heart of algebraic expressions by using the box volume problem as an example.

An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like our friends \( \ell \), \( w \), and \( h \) in the box volume problem), and operators (such as the plus and minus signs). The beautiful thing about these expressions is that they allow us to represent real-world quantities and relationships in a compact and generalizable form.

In the context of our box volume problem, the algebraic expressions help us translate the given relationships into a form we can work with mathematically. For instance, the formula \( \ell = h - 3 \) creates a connection between the length and height of the box, while \( w = h - 9 \) does the same for the width and height. Utilizing these expressions enables us to solve for unknown quantities by forming equations that depict physical scenarios, such as calculating the volume of a box.

Volume of a Box

When we talk about the volume of a box, we are referring to the amount of space enclosed within it. The concept is key when dealing with problems in math, especially those that involve three-dimensional objects. To calculate the volume of a rectangular box, you use the formula \( V = \ell \times w \times h \), where \( V \) denotes volume, \( \ell \) represents the length, \( w \) stands for the width, and \( h \) signifies the height.

In practical terms, think of it as stacking cubes. Each cube takes up a little bit of space, and by multiplying the number of cubes along each dimension we find out the total space the box occupies. The exercise in question has given a volume of 324 cubic inches, meaning our box would fit 324 of those little 1-inch cubes!

Crucially, our box has a specific relationship between its dimensions. The exercise tells us that the length and width are both less than the height by a certain measure. With this information, and knowing the total volume, we can work backwards to deduce each dimension, which is exactly what our problem is asking us to do.

Solving Cubic Equations

Solving cubic equations, such as the one found in the box volume problem, can be quite an adventure. A cubic equation is any equation with an unknown variable to the third power, often taking the form \( ax^3 + bx^2 + cx + d = 0 \). These equations require different tactics compared to linear or quadratic equations.

In our box volume example, the equation we need to solve is \( 324 = (h-3)(h-9)h \), a bona fide cubic equation with \( h \) as the variable. Breaking down such an equation might look daunting, but it typically involves methods such as factoring, using the Rational Root Theorem, or applying more advanced formulas.

The key to tackling cubic equations is to look for patterns and to simplify wherever possible. Once the cubic equation from our problem is solved for \( h \) (which may involve a few trials and errors or a systematic approach, depending on the complexity), we then use basic algebra to find the other dimensions. In many cases, including this one, the cubic equation can be solved by factoring out terms, or even better, by cleverly guessing and checking plausible values for \( h \) based on the context of the problem.