Maximize Prism Volume: A Math Challenge

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Hey math enthusiasts! Let's dive into a fun problem involving a rectangular prism and some clever optimization. We're going to use our math skills to figure out the maximum volume this prism can have, given some specific rules about its base and dimensions. Sounds like a blast, right?

The Problem Unpacked: Setting the Stage

Okay, so here’s the scoop. We're dealing with a rectangular prism, imagine a classic box shape. But this isn't just any box; it has some restrictions. First off, the area of its base can't be bigger than 27 square meters. Think of the base as the bottom of the box. Secondly, the width of this base is determined by the height of the prism. Specifically, the width is 9 meters less than the prism's height. And finally, the length of the base is 6 meters longer than its width. Got it? We need to find the maximum volume this box can achieve while sticking to these rules. This problem combines geometry with a touch of algebra – a perfect recipe for a satisfying mathematical challenge! Understanding the constraints is key; they're the boundaries within which we have to operate. It is important to know the volume of a rectangular prism is given by the formula: Volume = Length x Width x Height. Remember, the area of the base is Length x Width. This problem essentially asks us to find the specific dimensions (length, width, and height) that maximize the volume while adhering to the area and dimensional rules given.

Breaking Down the Constraints and Dimensions

Let's put the constraints and dimensions into a more useful format, where x represents the height. This way, everything can be calculated with a simple equation. Here's a quick recap of the constraints:

  • Base Area Constraint: The base area (Length × Width) must be ≤ 27 m².
  • Width vs. Height: Width = Height - 9 meters. If the height is 'x', then Width = x - 9.
  • Length vs. Width: Length = Width + 6 meters. Since Width is (x - 9), then Length = (x - 9) + 6 = x - 3.
  • Volume Formula: Volume = Length × Width × Height.

With these formulas, we can express all dimensions using the height, or 'x'. This is really helpful when we start building the formulas, because it makes everything easier to calculate. Remember, the height of the prism plays a crucial role in determining the length and width, thus greatly affecting the overall volume. Our target is to maximize the volume, which will depend on the value of x.

Formulating the Equations: The Math Mechanics

Now, let's get our hands dirty with some math! We're going to translate those wordy descriptions into mathematical equations. This is where the magic happens, and we start to see how everything fits together.

Expressing Length, Width, and Base Area

Using the relationships we found earlier, we can rewrite the length, width, and base area:

  • Length: L = x - 3
  • Width: W = x - 9
  • Base Area: A = Length × Width = (x - 3) × (x - 9)

The Volume Equation: Putting It All Together

Since Volume = Length × Width × Height and we've expressed everything in terms of the height 'x', we can write the volume equation as:

  • Volume (V): V = (x - 3) × (x - 9) × x

Now we have an equation for the volume that only depends on the height 'x'. This is a crucial step. This single formula now links all the constraints of the problem. However, there's another condition: the base area cannot exceed 27 square meters. This condition helps define the possible values of x.

Finding the Height: Unveiling the Limits

Now, we need to consider the base area constraint: The area of the base (Length × Width) must be less than or equal to 27 square meters. This constraint is critical because it puts a cap on how big our box can be. We have the equation for the base area: A = (x - 3) × (x - 9). This base area must be ≤ 27. So, let’s go ahead and solve that inequality to identify the permissible values for 'x' (the height).

Setting Up the Inequality

We will set up the following inequality:

(x - 3) × (x - 9) ≤ 27

Solving the Inequality

  1. Expand the Left Side: x² - 12x + 27 ≤ 27
  2. Simplify: x² - 12x ≤ 0
  3. Factor: x(x - 12) ≤ 0

This inequality gives us the critical points x = 0 and x = 12. Since we’re dealing with the dimensions of a prism, we can ignore the x = 0 result, since it would give us a volume of 0. The inequality is satisfied when x is between 0 and 12, inclusive. This means the height (x) must be within this range to meet the base area constraint. Therefore, we know that the height of the prism can't be greater than 12 meters. This range for height values will help us determine the possible volumes.

Maximizing the Volume: The Optimization Process

Alright, it's time to find that maximum volume. We know that the height, 'x', must be between 0 and 12. We can revisit the volume equation, V = (x - 3) × (x - 9) × x, to identify which height would give us the maximum volume.

Analyzing the Volume Equation

Looking at the volume equation, we're dealing with a cubic function. The shape of a cubic function has a unique curve, which rises and falls and can have local maxima and minima. However, because our constraint defines a range for x, we will be looking for a local maximum within that range.

Considering Possible Solutions

We know that we must consider the boundaries, and also the internal critical points if any. Let's analyze the critical points in detail. Our height is subject to the constraints we previously computed, which gives us an effective range of 0 < x ≤ 12. To truly get a sense of how the volume changes with the height, you could also substitute some values for x into the volume equation. Doing so will help us to understand where the volume reaches a maximum.

Finding the Optimal Height

Since the width, x-9, has to be greater than 0, we can also say that x > 9. This means that x must be somewhere between 9 and 12. The best way to identify the max volume is to plug in the highest value of x, which is 12, to see what value of V we get. So,

  • V = (12 - 3) × (12 - 9) × 12 = 324

Calculating the Dimensions and Final Volume

With all the information we have, let's calculate the dimensions and final volume. We have the height, so now we can find the width and the length.

  • Height (x): 12 meters
  • Width: 12 - 9 = 3 meters
  • Length: 12 - 3 = 9 meters
  • Maximum Volume (V): 9 × 3 × 12 = 324 cubic meters

So, the maximum volume of the rectangular prism, given all the restrictions, is 324 cubic meters. Pretty cool, huh? The optimal height is 12 meters, which gives us these dimensions: 9 meters for length and 3 meters for width.

Conclusion: A Mathematical Victory

We did it, guys! We successfully tackled this rectangular prism problem and found the maximum volume. We combined geometric principles, algebraic manipulation, and inequalities to reach our answer. Remember, the trick was understanding the constraints, setting up the equations correctly, and working through the math systematically. Each step, from understanding the base area constraint to figuring out the height, played a vital role in finding that maximum volume. This problem really highlights how different math concepts can work together to solve a real-world problem. And that's exactly what makes mathematics so awesome!