Problem 48 - Calculus for Biology and Medicine

Question

Let $$f(x, y)=x y, \quad(x, y) \in \mathbf{R}^{2}$$ with constraint function \(y-x^{2}=0\). (a) Use Lagrange multipliers to find candidates for local extrema. (b) Use the constraint \(y-x^{2}=0\) to reduce \(f(x, y)\) to a singlevariable function, and then use this function to show that \(f(x, y)\) has no local extrema on the constraint curve.

Step-by-Step Solution

Answer

The function has no local extrema on the constraint curve.

Step 1: Set up the Lagrange function

To use the method of Lagrange multipliers, we set up the Lagrange function \( \mathcal{L}(x, y, \lambda) = f(x, y) + \lambda g(x, y) \), where \( g(x, y) = y - x^2 = 0 \) is the constraint. Thus, \[ \mathcal{L}(x, y, \lambda) = xy + \lambda(y - x^2). \]

Step 2: Find the gradients

We calculate the gradients of the Lagrange function with respect to \(x, y,\) and \(\lambda\):- \( \frac{\partial \mathcal{L}}{\partial x} = y - 2\lambda x = 0 \)- \( \frac{\partial \mathcal{L}}{\partial y} = x + \lambda = 0 \)- \( \frac{\partial \mathcal{L}}{\partial \lambda} = g(x, y) = y - x^2 = 0 \)

Step 3: Solve the system of equations

We solve the system of equations obtained in Step 2:1. From \( \frac{\partial \mathcal{L}}{\partial x} = 0 \), we have \( y = 2\lambda x \).2. From \( \frac{\partial \mathcal{L}}{\partial y} = 0 \), we have \( x + \lambda = 0 \), hence \( \lambda = -x \).3. Substitute \( \lambda = -x \) into \( y = 2\lambda x \) to get \( y = -2x^2 \).Since \( y = x^2 \) from the constraint, setting \( x^2 = -2x^2 \) implies \( 3x^2 = 0 \), leading to \( x = 0 \). Thus, \( y = 0 \) as well.

Step 4: Verify critical points

The only critical point is \( (0, 0) \). Substitute back into the function to find that \( f(0, 0) = 0 \).

Step 5: Reduce to a single-variable function

Using the constraint \( y = x^2 \), substitute into \( f(x, y) = xy \) to get the single-variable function \( f(x) = x \cdot x^2 = x^3 \).

Step 6: Analyze the single-variable function

The function \( f(x) = x^3 \) is a cubic function with derivative \( f'(x) = 3x^2 \). The derivative is zero only at \(x = 0\), indicating a stationary point. However, because the second derivative \( f''(x) = 6x \) is also zero here, we must inspect further.

Step 7: Investigate behavior of \( f(x) = x^3 \)

Since \( x^3 \) does not change sign around \( x = 0 \), it does not have any local extrema. Its local behavior resembles that of its graph, indicating no local minima or maxima.

Key Concepts

Constraint Function

The concept of a constraint function plays a crucial role when you're dealing with optimization problems using methods like Lagrange multipliers. In the given problem, the constraint function is defined as \( g(x, y) = y - x^2 = 0 \). This condition represents a curve on the \(xy\)-plane known as the constraint curve. It essentially restricts the values that \(x\) and \(y\) can take while still being solutions to the problem.

When you have a constraint function, you're not just trying to find the maximum or minimum of a function like \(f(x, y) = xy\), but you're doing so under specific conditions. Think of it as looking for the top of a hill not anywhere on the landscape, but on a defined path or line. Constraint functions tell us exactly where to look.

In this problem, we used the constraint \( y = x^2 \) to simplify and direct our search for local extrema, by reducing the function \(f(x, y)\) to a function in a single variable. This simplifies the problem significantly, by focusing the exploration of extrema to a specific pathway.

Local Extrema

Local extrema refer to the points where a function reaches local maxima or minima. In the context of the original exercise, we are looking for points on the constraint curve where the function \(f(x, y) = xy\) could possibly be the highest (maximum) or lowest (minimum) value under given conditions.

Using the method of Lagrange multipliers, we calculate the gradients of both the function and the constraint and set them equal to each other. This helps in determining the critical points, which are candidates for local extrema. In the context of the problem, we found \((0,0)\) to be the only critical point.

However, upon further investigation and reducing the function to a single-variable form, we find that there are no local extrema. Specifically, even though \( (0, 0) \) is a stationary point, the behavior of the function does not suggest it is an extremum as \(f(x) = x^3\) continues smoothly without reaching a peak or valley at this point.

Single-Variable Function

When dealing with a problem like this, reducing a complex multi-variable function to a simpler single-variable function can greatly ease the process of analyzing the function's behavior. With the constraint \(y = x^2\), the function \(f(x, y) = xy\) can be transformed into a function of a single variable: \(f(x) = x^3\).

The benefit of this transformation is that it allows us to use more straightforward calculus methods to analyze the function. After converting \(f(x, y)\) into \(f(x)\), we computed its derivative \(f'(x) = 3x^2\), which helps identify critical points. In this case, \(f'(x) = 0\) only at \(x = 0\), showing \(x = 0\) as a stationary point.

Further inspection of the derivative and the second derivative \(f''(x) = 6x\) demonstrates that \(x^3\) doesn't really have a change in direction that would suggest a local maximum or minimum at this point. By looking at its graph, you can see that \(x^3\) crosses \(x = 0\) smoothly, confirming no local extrema exist for this single-variable function.