Problem 51 - College Algebra and Calculus: An Applied Approach

Question

Dimensions of a Terrarium A rectangular terrarium with a square cross section has a combined length and girth (perimeter of a cross section) of 108 inches (see figure). Find the dimensions of the terrarium, given that the volume is 11,664 cubic inches.

Step-by-Step Solution

Answer

Taking L = 36 and 's' = 18 inches, the dimensions are Length = 36 inches and Width (or height) = 18 inches.

Step 1: Develop the First Equation

From the problem, we know that the combined length (L) and girth (4 * side of the square cross section, 4s) is 108 inches. This gives us the equation L + 4s = 108.

Step 2: Develop the Second Equation

The volume (V) of the terrarium = L * s^2 (since it has a square cross-section), which is given as 11,664 cubic inches. This gives us the equation L * s^2 = 11664.

Step 3: Solve For 's' From First Equation

Rearrange the first equation, L + 4s = 108, to find 's'. We get s = (108 - L) / 4

Step 4: Substitute 's' in the Second Equation

Substitute the value of 's' in second equation, we get: L * ((108 - L) / 4)^2 = 11664.

Step 5: Solve the Quadratic Equation

Solving the above equation will give us the value of L. Once we have the value of L, we can substitute it in equation of 's' to get the side of the square cross-section.

Key Concepts

Rectangular Terrarium Dimensions

Understanding the dimensions of a rectangular terrarium with a square cross-section involves deciphering the relationship between the terrarium’s length, width, and height. The essential aspect to note is that the width and height are equal, which is typical for a square cross-section. In this specific scenario, the information presents a combined measurement of length and girth at 108 inches. Hence, if we let the side length of the square cross-section be 's', then the girth, which is the perimeter of the square base, would be 4s. To find an individual dimension, we establish the equation based on given parameters:

\[\begin{equation} L + 4s = 108 \end{equation}\] where L represents the length, and s represents the side of the square cross-section. When solving for the dimensions, this equation will prove central to determining the individual measurements of the terrarium.

Volume of a Rectangular Terrarium

The volume of a terrarium is the amount of space inside it, and is crucial for determining how much soil, water, and living organisms it can contain. For a rectangular prism or terrarium with a square cross-section, the volume is computed by multiplying the length by the square of the side of the cross-section. The equation given for our terrarium with volume 11,664 cubic inches is:

\[\begin{equation} L \times s^2 = 11,664 \end{equation}\] This formula is vital for connecting the dimensions with the capacity of the terrarium. Knowing that the volume is fixed allows us to find a proportional relationship between length and the side of the square cross-section, which helps to resolve the measurements needed to construct the terrarium.

Solving Quadratic Equations

The terrarium problem ultimately boils down to solving a quadratic equation to find the terrarium’s length. Once that is achieved, the side dimension follows. A quadratic equation is a second-degree polynomial equation, typically taking the form \[\begin{equation} ax^2 + bx + c = 0 \end{equation}\] The technique used to solve the equation varies depending on the specific equation structure, yet common methods include factoring, completing the square, or utilizing the quadratic formula. When substituting the value of 's' from equation \[\begin{equation} L + 4s = 108 \end{equation}\] into equation \[\begin{equation} L \times s^2 = 11,664 \end{equation}\], we end up with a quadratic equation in terms of L. Solving this equation will yield the length of the terrarium, and then by back-substituting the found value of L into the initial equation, we can have all dimensions of the terrarium.
The process of simplifying and solving quadratic equations is a foundation of algebra that extends into calculus and beyond. Understanding how to manipulate and solve these equations not only aids in resolving practical problems, like our terrarium dimensions issue, but also paves the way for analyzing more complex mathematical concepts.