Maximize & Minimize $f(x, Y) = E^{xy}$ With $x^3 + Y^3 = 128$

Hey math enthusiasts! Today, we're diving into a classic optimization problem. Our mission, should we choose to accept it, is to find the maximum and minimum values of the function $f(x, y) = e^{xy}$, but with a twist: we have the constraint $x^3 + y^3 = 128$. This means we're not just looking at any old $x$ and $y$; they have to play by the rules of our equation. It's like a treasure hunt, and our constraint is the map guiding us to the hidden gems – the maximum and minimum values of our function. Let's break this down step by step and get to the good stuff. We'll use the method of Lagrange multipliers to tackle this problem. This method is a real lifesaver when you need to optimize a function subject to a constraint. Get ready to flex those math muscles – this is going to be fun!

Understanding the Problem and Lagrange Multipliers

Alright, before we jump into the nitty-gritty, let's make sure we're all on the same page. We're trying to find the highest and lowest points (the maximum and minimum values) that our function $f(x, y) = e^{xy}$ can reach. Think of it like a mountain range; our function is the elevation, and we're looking for the peaks and valleys. But here’s the catch: we can only explore a specific path, defined by the constraint $x^3 + y^3 = 128$. This constraint is like a winding road on our mountain range – we can only move along this road. We use the Lagrange multipliers to navigate the winding road. The core idea behind Lagrange multipliers is that at the maximum or minimum points, the gradient of our function $f$ is parallel to the gradient of our constraint $g(x, y) = x^3 + y^3 - 128 = 0$. This means their directions are the same (or opposite), and that’s super helpful! Mathematically, this gives us the equation $abla f = \lambda abla g$, where $\lambda$ (lambda) is our Lagrange multiplier.

So, what does this all mean in practice? We need to calculate the gradients of both our function and our constraint. The gradient is like a vector that points in the direction of the steepest increase of a function. The gradient of $f(x, y) = e^{xy}$ is $abla f = <y e^{xy}, x e^{xy}>$. The gradient of $g(x, y) = x^3 + y^3 - 128$ is $abla g = <3x^2, 3y^2>$. Now, we set up our Lagrange multiplier equations. These equations will help us find the values of $x$, $y$, and $\lambda$ that satisfy both the function and the constraint. Remember, at the heart of this method, we are using the constraint equation along with the gradient equality to find our solutions. The constraint equation itself is a key component, ensuring that the found values adhere to the condition provided.

Now, we'll set up the equations. The gradient equality gives us: $y e^{xy} = \lambda (3x^2)$ and $x e^{xy} = \lambda (3y^2)$. We also have the constraint equation: $x^3 + y^3 = 128$. Now, let's get solving!

Setting Up the Equations and Solving for Critical Points

Alright, guys, time to get our hands dirty with some equations! Based on the Lagrange multiplier method, we've got the following system of equations to solve:

1. $y e^{xy} = 3\lambda x^2$
2. $x e^{xy} = 3\lambda y^2$
3. $x^3 + y^3 = 128$

Our goal is to find the values of $x$ and $y$ that satisfy all these equations simultaneously. This is where the magic happens – the algebra that brings us closer to those maximum and minimum values. First, let's divide equation (1) by equation (2), assuming neither $x$ nor $y$ is zero (we'll address this assumption later). This gives us:

$\frac{y e^{xy}}{x e^{xy}} = \frac{3\lambda x^2}{3\lambda y^2}$

Which simplifies to:

$\frac{y}{x} = \frac{x^2}{y^2}$

Cross-multiplying, we get $y^3 = x^3$. Taking the cube root of both sides gives us $y = x$. Now, we have a critical piece of the puzzle: a direct relationship between $x$ and $y$. Next, we'll substitute $y = x$ into our constraint equation (3): $x^3 + y^3 = 128$, becoming $x^3 + x^3 = 128$, or $2x^3 = 128$. Dividing both sides by 2, we get $x^3 = 64$. Taking the cube root of both sides, we find $x = 4$. Since $y = x$, it immediately follows that $y = 4$. So, we've found our first critical point: $(4, 4)$. Now, let's evaluate our function at this point: $f(4, 4) = e^{(4)(4)} = e^{16}$. But wait, is this a maximum or a minimum? We'll figure that out soon!

Let's address the assumption we made earlier that neither $x$ nor $y$ is zero. If either $x$ or $y$ were zero, it would imply, according to the constraint $x^3 + y^3 = 128$, that the other variable would be the cube root of 128, which is not zero. So, our assumption was valid for our calculations. Now, let's keep going to find other potential critical points and determine the nature of the point we've found.

Determining Maximum and Minimum Values

Alright, we've got a critical point $(4, 4)$ and the value of our function at this point, $f(4, 4) = e^{16}$. But is this a maximum, a minimum, or something else entirely? We need to investigate further to determine the nature of this point. One way to do this is to consider the behavior of our function along the constraint. In this case, since we only have one critical point, and the constraint is continuous, we can analyze the behavior of the function around this point. Remember, our function is $f(x, y) = e^{xy}$, and it's an exponential function, which means it will always output a positive value. Also, because $x^3 + y^3 = 128$, we are dealing with a bounded region, which means our function is guaranteed to have both a maximum and a minimum value within this region. The function $e^{xy}$ is always positive, and the exponent $xy$ determines the magnitude. The critical point we found, $(4, 4)$, leads to $xy = 16$. The value of $e^{16}$ is extremely large; hence, we can infer that this is our maximum value. To confirm and find the minimum, we must consider the endpoints of the constraint. Let's analyze. If $x$ is a large positive number, then $y$ must be a small positive number to satisfy the constraint. Conversely, if $x$ is a small positive number, then $y$ is a large positive number. Thus, the product $xy$ would be smaller than 16, resulting in a value of $f(x, y)$ that is less than $e^{16}$. This means there must be a minimum value somewhere on the constraint, but it's not straightforward to pinpoint another critical point using the Lagrange multiplier method alone in this case. In such scenarios, the endpoints of the domain are often crucial in determining the absolute maximum and minimum. Given the nature of $x^3 + y^3 = 128$, the variables $x$ and $y$ must be positive, and we know our critical point. When $x$ is close to 0 or $y$ is close to 0, $xy$ gets close to 0, so the minimum value of $f(x, y)$ approaches $e^0 = 1$ when either $x$ or $y$ approaches 0. Therefore, the minimum value would be approaching 1 as $x$ or $y$ approaches 0, and we take our value to be $1.0000$ (rounded to 4 decimal places), and our maximum value is $e^{16}$. To calculate $e^{16}$ to four decimal places, we get $8886110.5205$.

• Maximum Value: $f(4, 4) = e^{16} \approx 8886110.5205$
• Minimum Value: The minimum value is reached as $x$ or $y$ approaches 0, thus $f(x, y) = 1.0000$

So there you have it, folks! We've successfully navigated the twists and turns of our optimization problem, using Lagrange multipliers to find the maximum and minimum values of our function. Pretty neat, huh?

Conclusion

To sum it all up, we utilized the Lagrange multiplier method to find the maximum and minimum values of the function $f(x, y) = e^{xy}$, subject to the constraint $x^3 + y^3 = 128$. We identified a critical point, $(4, 4)$, which gave us the maximum value of the function as approximately 8886110.5205. The constraint guided our exploration, showing that while our maximum value is exceptionally large, the minimum value approaches 1 as either $x$ or $y$ gets close to 0. Understanding the interplay between a function and its constraints is a cornerstone of calculus and optimization, and with each problem, we sharpen our problem-solving skills and deepen our mathematical understanding. Keep practicing, keep exploring, and keep the mathematical journey alive! If you've got any questions or want to explore more problems, feel free to ask. Keep the math spirit alive, guys!