Maximum And Minimum Volume Calculation For A Rectangular Box

Hey guys! Today, we're diving into a super interesting problem from mathematics: finding the maximum and minimum volumes of a rectangular box. We're given the volume equation, and our mission is to figure out those special x values that make the volume either the biggest or the smallest it can be. Sounds like a fun challenge, right? Let's break it down and see how we can crack this!

Understanding the Problem

First things first, let's make sure we're all on the same page. We have a rectangular box, and its volume, V, is described by the equation V = x³ - 8x² + 5x + 50. Our main goal here is to calculate the values of x for which this volume, V, reaches its maximum or minimum. In simpler terms, we want to find the x values that make the box either as big as possible or as small as possible. This involves a bit of calculus, but don't worry, we'll take it step by step. We'll be using derivatives to find the critical points, which are the potential locations of our maximum and minimum volumes. These points occur where the rate of change of the volume (V) with respect to x is zero or undefined. Once we find these critical points, we'll use the second derivative test to determine whether they correspond to a maximum or a minimum volume. This test tells us about the concavity of the volume function at these points. If the second derivative is positive, we have a local minimum, and if it's negative, we have a local maximum. The key here is understanding that the derivative gives us the slope of the tangent line to the volume function. At maximum and minimum points, this slope is zero, meaning the tangent line is horizontal. This is why setting the derivative equal to zero helps us find these points. Remember, we're dealing with a real-world scenario – the volume of a box. So, our x values need to make sense in this context. We can't have negative dimensions, so we'll need to consider the physical constraints of the problem when interpreting our results. We'll also need to think about whether our solutions truly represent a maximum or minimum volume in the practical sense, perhaps by considering the domain of x and the behavior of the volume function at the endpoints of that domain.

Step 1: Finding the First Derivative

The first thing we need to do is find the first derivative of our volume equation. This will help us understand how the volume changes as x changes. Remember, the derivative tells us the slope of the curve at any point. So, let's differentiate V with respect to x. To find the first derivative, we'll apply the power rule, which states that the derivative of xⁿ is nxⁿ⁻¹. So, we'll apply this rule to each term in our volume equation, V = x³ - 8x² + 5x + 50. Differentiating x³ gives us 3x², differentiating -8x² gives us -16x, differentiating 5x gives us 5, and the derivative of the constant 50 is 0. Combining these, we get the first derivative, dV/dx = 3x² - 16x + 5. This new equation represents the rate of change of the volume with respect to x. It's a quadratic equation, which means it represents a parabola. The roots of this equation will give us the x values where the volume's rate of change is zero, which are our potential maximum or minimum points. We'll use this derivative in the next step to find the critical points where the volume could be at its maximum or minimum. Understanding the derivative is crucial here because it gives us the tool to analyze how the volume function behaves. It tells us where the function is increasing, decreasing, and where it has potential turning points. The first derivative is essentially a roadmap that guides us to the points of interest in our volume function. It's the key to unlocking the maximum and minimum values we're searching for. So, make sure you're comfortable with the concept of derivatives and how they relate to the rate of change before moving on. Got it? Great, let's move on to the next step!

Step 2: Finding Critical Points

Now that we have the first derivative, 3x² - 16x + 5, we need to find the critical points. These are the points where the derivative equals zero, meaning the slope of the tangent line is horizontal. These points are crucial because they indicate potential maximum or minimum volumes. To find these points, we set the first derivative equal to zero and solve for x: 3x² - 16x + 5 = 0. This is a quadratic equation, and we can solve it by factoring, using the quadratic formula, or completing the square. In this case, factoring works nicely. We're looking for two numbers that multiply to 15 (3 * 5) and add up to -16. Those numbers are -15 and -1. So, we can rewrite the equation as 3x² - 15x - x + 5 = 0. Now, we can factor by grouping: 3x(x - 5) - 1(x - 5) = 0. This gives us (x - 5)(3x - 1) = 0. Setting each factor equal to zero gives us two possible solutions: x - 5 = 0, which means x = 5, and 3x - 1 = 0, which means x = 1/3. These are our critical points. They're the x values where the volume function could have a maximum or minimum. But we don't know for sure yet! We need to use the second derivative test to determine whether these points correspond to a maximum, a minimum, or neither. Remember, critical points are like intersections on our roadmap. They tell us where we might need to make a turn. But we need more information to know whether that turn will lead us uphill (to a maximum) or downhill (to a minimum). That's where the second derivative comes in. It's like a signpost that tells us which direction we're heading.

Step 3: Finding the Second Derivative

The second derivative will help us determine whether our critical points are maximums or minimums. It tells us about the concavity of the volume function. To find the second derivative, we simply differentiate the first derivative again. Our first derivative was dV/dx = 3x² - 16x + 5. Applying the power rule again, we differentiate each term. The derivative of 3x² is 6x, the derivative of -16x is -16, and the derivative of the constant 5 is 0. So, our second derivative is d²V/dx² = 6x - 16. This equation is a linear function, and its value at our critical points will tell us about the concavity of the volume function at those points. A positive second derivative means the function is concave up (like a smile), indicating a local minimum. A negative second derivative means the function is concave down (like a frown), indicating a local maximum. The second derivative is like a curvature detector. It tells us how the slope of the volume function is changing. If the slope is increasing (positive second derivative), we're on the upward curve of a minimum. If the slope is decreasing (negative second derivative), we're on the downward curve of a maximum. This is a crucial step in identifying whether our critical points are indeed maximums or minimums. Without the second derivative, we'd be guessing! So, let's use this powerful tool to analyze our critical points and find those maximum and minimum volumes.

Step 4: Applying the Second Derivative Test

Now, let's put our second derivative to work! We'll use it to test our critical points, x = 5 and x = 1/3, and determine whether they correspond to a maximum or minimum volume. Remember, our second derivative is d²V/dx² = 6x - 16. First, let's plug in x = 5 into the second derivative: d²V/dx² = 6(5) - 16 = 30 - 16 = 14. Since 14 is positive, this means the volume function is concave up at x = 5, indicating a local minimum. So, at x = 5, the volume is at a minimum. Next, let's plug in x = 1/3 into the second derivative: d²V/dx² = 6(1/3) - 16 = 2 - 16 = -14. Since -14 is negative, this means the volume function is concave down at x = 1/3, indicating a local maximum. So, at x = 1/3, the volume is at a maximum. The second derivative test is like a detective that helps us solve the mystery of maximums and minimums. By evaluating the sign of the second derivative at our critical points, we can confidently classify them as either local maximums or local minimums. This test is a powerful tool in calculus, and it's essential for understanding the behavior of functions. Now that we've applied the second derivative test, we know that x = 5 corresponds to a minimum volume and x = 1/3 corresponds to a maximum volume. But we're not quite done yet! We still need to think about the practical implications of these results and consider the physical constraints of our problem. Let's move on to the final step to wrap things up.

Step 5: Interpreting the Results

We've found that x = 5 corresponds to a local minimum volume and x = 1/3 corresponds to a local maximum volume. But what does this really mean in the context of our rectangular box? It's super important to interpret these results in a way that makes sense for the real world. Remember, x represents a dimension of the box, so it can't be negative. Also, the volume V = x³ - 8x² + 5x + 50 needs to be a real number. Let's think about the minimum volume first. At x = 5, we have a local minimum. To find the actual minimum volume, we plug x = 5 back into the original volume equation: V = (5)³ - 8(5)² + 5(5) + 50 = 125 - 200 + 25 + 50 = 0. So, the minimum volume is 0. This makes sense because a box can't have a negative volume. Now, let's think about the maximum volume at x = 1/3. Plugging this value into the volume equation gives us: V = (1/3)³ - 8(1/3)² + 5(1/3) + 50 = 1/27 - 8/9 + 5/3 + 50. To simplify this, we need a common denominator, which is 27: V = 1/27 - 24/27 + 45/27 + 1350/27 = 1372/27 ≈ 50.81. So, the maximum volume is approximately 50.81 cubic units. Interpreting our results is crucial because it connects the mathematical solution to the physical reality of the problem. We need to ask ourselves if our answers make sense in the real world. Can a box have a negative dimension? Can a volume be negative? By considering these questions, we can ensure that our solution is not only mathematically correct but also practically meaningful. In this case, we've found that the minimum volume is 0, which is a valid physical limit, and the maximum volume is approximately 50.81 cubic units, which is a reasonable value. So, we've successfully found the x values that give us the maximum and minimum volumes of our rectangular box! Great job, guys! You've tackled a challenging problem and come out on top!

Conclusion

So, there you have it! We've successfully navigated the world of calculus to find the values of x that maximize and minimize the volume of our rectangular box. We found that the minimum volume occurs at x = 5, and the maximum volume occurs at x = 1/3. Remember, the key to solving these types of problems is to break them down into smaller, manageable steps. We started by finding the first derivative, which told us about the rate of change of the volume. Then, we found the critical points by setting the first derivative equal to zero. Next, we used the second derivative test to determine whether these points were maximums or minimums. Finally, we interpreted our results in the context of the problem, making sure they made sense in the real world. Calculus can seem intimidating at first, but with practice and a clear understanding of the concepts, you can solve even the most challenging problems. Keep practicing, keep exploring, and most importantly, keep having fun with math! You've got this!