Math Problem: Box Volume & Dimensions

Hey math enthusiasts! Ever found yourself staring at a shipping box, wondering about the math behind its dimensions? Today, we're diving into a classic problem that involves calculating the volume of a rectangular prism, a concept fundamental to geometry and everyday life. Imagine Jordan, who just made an online purchase and is eagerly tracking its arrival. The shipping details are a bit puzzling: the item is set to arrive in a box that's 24 inches long, and its total volume is a hefty 2,880 cubic inches. But there's a twist! The width of this box is directly related to its height – specifically, the width is seven inches less than the height. This scenario presents us with a fantastic opportunity to flex our mathematical muscles and figure out the exact dimensions of this shipping box. We'll be using our knowledge of algebraic equations and the formula for the volume of a rectangular prism to crack this case. So, grab your calculators, and let's get started on unraveling this dimension mystery!

Understanding the Volume of a Rectangular Prism

At the heart of this problem lies the concept of the volume of a rectangular prism. Think of a rectangular prism as a fancy word for a box – something with length, width, and height. The volume of such a shape is essentially how much space it occupies. We calculate this by multiplying its three dimensions together: Volume = Length × Width × Height. This formula is incredibly versatile and is used in everything from calculating the amount of liquid a container can hold to determining the amount of material needed for construction. In Jordan's case, we know the total volume (2,880 cubic inches) and the length (24 inches). This means we have a good starting point to find the missing width and height. The key here is to recognize that even though we have one unknown dimension (or rather, two related unknowns), the provided volume acts as our anchor. We'll be setting up an equation where the product of the length, width, and height equals the given volume. This is where the real problem-solving begins, as we'll need to translate the word problem into a mathematical expression that we can solve. The more you understand this fundamental formula, the easier it will be to tackle more complex geometry problems. It’s a cornerstone of spatial reasoning, allowing us to quantify three-dimensional objects in a way that’s both practical and elegant. So, before we jump into solving Jordan's specific dilemma, make sure you're comfortable with the idea that Volume = L × W × H is your best friend when dealing with boxes and similar shapes. This foundational knowledge will empower you to break down any problem involving three-dimensional spaces into manageable steps. We’re not just finding numbers here; we’re learning to see the world in terms of dimensions and capacities, which is a surprisingly useful skill!

Setting Up the Algebraic Equation

Now, let's get down to setting up the algebraic equation needed to solve Jordan's shipping box mystery. We know the formula for the volume of a rectangular prism: $V = L imes W imes H$. We are given the volume ($V = 2,880$ cubic inches) and the length ($L = 24$ inches). The critical piece of information that links the width ($W$) and height ($H$) is that the width is seven inches less than the height. We can express this relationship mathematically. If we let the height be represented by the variable $h$, then the width, being seven inches less, can be represented as $h - 7$. So, our equation starts to take shape. We can substitute the known values and these variable expressions into the volume formula:

$2880 = 24 imes (h - 7) imes h$

This equation now contains only one unknown variable, $h$ (the height), which is exactly what we want. Our goal is to isolate $h$ and solve for it. Before we dive into solving this quadratic equation, let's simplify it a bit. We can divide both sides of the equation by the length (24) to get:

rac{2880}{24} = (h - 7) imes h

$120 = h(h - 7)$

Now, we can distribute the $h$ on the right side:

$120 = h^2 - 7h$

To solve this quadratic equation, we need to set it equal to zero by moving the 120 to the other side:

$h^2 - 7h - 120 = 0$

This is a standard quadratic equation in the form $ax^2 + bx + c = 0$, where $a=1$, $b=-7$, and $c=-120$. We now have a clear path to finding the height, and subsequently the width, of the box. The ability to translate word problems into solvable algebraic equations is a cornerstone of applied mathematics, and this step is crucial in bridging the gap between the narrative and the numerical solution. It requires careful reading, precise variable assignment, and a solid understanding of how mathematical relationships can be represented symbolically. We've successfully transformed Jordan's shipping conundrum into a form that's ready for us to solve algebraically, paving the way for the next stage of our investigation into the box's dimensions.

Solving for the Height and Width

With our algebraic equation neatly set up as $h^2 - 7h - 120 = 0$, we can now proceed to solve for the height ($h$). There are a couple of common methods to solve quadratic equations: factoring or using the quadratic formula. Let's try factoring first, as it's often quicker if the equation is factorable. We're looking for two numbers that multiply to -120 and add up to -7. After a bit of thought, we can identify these numbers as -15 and +8. They multiply to -120 (-15 × 8 = -120) and add up to -7 (-15 + 8 = -7). So, we can factor the equation as:

$(h - 15)(h + 8) = 0$

For this product to be zero, at least one of the factors must be zero. This gives us two possible solutions for $h$:

$h - 15 = 0 ightarrow h = 15$

$h + 8 = 0 ightarrow h = -8$

In the context of a physical box, a negative height doesn't make sense. Therefore, we discard the solution $h = -8$. The height of the box must be 15 inches. Now that we have the height, we can easily find the width using the relationship we established earlier: width ($W$) is seven inches less than the height.

$W = h - 7$

$W = 15 - 7$

$W = 8$ inches

So, the width of the box is 8 inches. We have successfully found both the height and the width of Jordan's shipping box. It's always a good practice to double-check our answers. Let's plug these dimensions back into the volume formula along with the given length:

$V = L imes W imes H$

$V = 24 imes 8 imes 15$

$V = 192 imes 15$

$V = 2880$

This matches the given volume of 2,880 cubic inches, confirming that our calculations for height and width are correct. This process highlights how algebraic techniques can be used to solve practical, real-world problems involving geometry. The problem required us to not only understand the formula for the volume of a rectangular prism but also to translate a word description into an algebraic equation and then solve that equation accurately, considering the physical constraints of the situation. It’s a neat demonstration of how math connects to the tangible world around us, from online shopping to architectural designs. The ability to solve such problems demonstrates a strong grasp of mathematical principles and their applications.

The Final Dimensions and Conclusion

We've successfully navigated the mathematical landscape to determine the final dimensions of Jordan's shipping box. Recall that the problem stated the box was 24 inches long with a volume of 2,880 cubic inches, and that the width was seven inches less than the height. Through careful application of the volume of a rectangular prism formula and algebraic problem-solving, we've determined that the height of the box is 15 inches and the width of the box is 8 inches. So, the complete dimensions of the box are 24 inches (length) × 8 inches (width) × 15 inches (height). We confirmed this by multiplying these dimensions: $24 imes 8 imes 15 = 2880$ cubic inches, which perfectly matches the given volume. This problem serves as an excellent example of how mathematics, particularly algebra and geometry, is woven into our everyday experiences. Whether it's understanding shipping logistics, planning packaging, or even designing spaces, the principles we've applied here are fundamental. It’s a reminder that math isn't just confined to textbooks; it's a practical tool that helps us make sense of and interact with the world around us. By breaking down the problem into smaller, manageable steps – understanding the core formula, translating the word problem into an equation, and then solving that equation – we were able to arrive at the correct answer. This methodical approach is key to tackling any complex problem, mathematical or otherwise.

For those interested in further exploring geometric calculations and their real-world applications, resources like Khan Academy offer excellent tutorials and practice problems on topics such as volume, surface area, and other geometric principles. They provide a wealth of information that can deepen your understanding and hone your problem-solving skills in a clear and accessible manner.