Question

Suppose that \(f(x, y)=x y .\) Find the maximum value for \(f\) if \(x\) and \(y\) are constrained to sum to \(1 .\) Solve this problem in two ways: by substitution and by using the Lagrange multiplier method.

Short answer

Answer: The maximum value of the function is \(\frac{1}{4}\) when \(x = \frac{1}{2}\) and \(y = \frac{1}{2}\).

Step-by-step solution

Substitution Method

First, let's solve the constraint equation \(x + y = 1\) for one of the variables, say \(y\). We get \(y = 1 - x\). Now substitute this expression for \(y\) in the function \(f(x, y)\): $$f(x, 1 - x) = x(1 - x) = x - x^2$$ Now, we'll find the maximum value of this single-variable function. To do so, we can compute the first derivative and find the critical points by setting the derivative equal to zero. $$\frac{d}{dx}(x - x^2) = 1 - 2x$$ Setting the first derivative equal to zero gives: $$1 - 2x = 0 \Rightarrow x = \frac{1}{2}$$ Since there's only one critical point, it's either a minimum or maximum. Check the second derivative to determine which: $$\frac{d^2}{dx^2}(x - x^2) = -2$$ The second derivative is negative, indicating that the critical point is a maximum. Now, substitute the maximum \(x\) back into the constraint equation to find the corresponding \(y\): $$y = 1 - \frac{1}{2} = \frac{1}{2}$$ With \(x = \frac{1}{2}\) and \(y = \frac{1}{2}\), the maximum value of the function is: $$f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{4}$$

Lagrange Multiplier Method

Now, let's solve the same problem using the Lagrange multiplier method. Define a new function \(F(x, y, \lambda) = xy - \lambda(x + y - 1)\), where \(\lambda\) is the Lagrange multiplier. Compute the partial derivatives of this function with respect to \(x, y,\) and \(\lambda\). The following system of equations represents the critical points of \(F\): $$\frac{\partial F}{\partial x} = y - \lambda = 0 \hspace{1cm} (1)$$ $$\frac{\partial F}{\partial y} = x - \lambda = 0 \hspace{1cm} (2)$$ $$\frac{\partial F}{\partial \lambda} = x + y - 1 = 0 \hspace{1cm} (3)$$ Solving equation (1) for \(\lambda\) gives \(\lambda = y\). Substituting that into equation (2) yields: $$x = y$$ Now, substitute \(x = y\) into equation (3) to obtain: $$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$$ Since \(x = y\), we have \(y = \frac{1}{2}\). So, our critical point is at \((\frac{1}{2}, \frac{1}{2})\). Due to the constraint, this is the only feasible critical point, and the maximum value of the function is the same as in the substitution method: $$f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{4}$$ In both methods, we find that the maximum value of \(f(x, y) = xy\) under the constraint \(x + y = 1\) is \(\frac{1}{4}\), achieved when \(x = \frac{1}{2}\) and \(y = \frac{1}{2}\).

Key concepts

Constrained Optimization

Constrained optimization is a cornerstone concept in calculus and economics, utilized to find the maximum or minimum values of functions under given constraints. Imagine you're trying to maximize your score on a test, but you must study within a limited time frame. In mathematical terms, the score is the function to maximize, and the time frame is the constraint.

Specifically, it involves a target function, commonly denoted as \(f(x,y)\), and constraints given by other functions, such as \(g(x,y) = c\). The goal is to find the point \((x, y)\) which maximizes or minimizes \(f(x,y)\) while satisfying the constraint \(g(x,y) = c\). This is a fundamental technique used across various scientific and engineering fields where resources are limited, conditions must be met, or certain balances must be maintained.

Substitution Method

The substitution method simplifies a constrained optimization problem by reducing the number of independent variables. This is done by expressing one variable in terms of another using the constraint equation. Following our test score analogy, it's like figuring out how much time to allot to each subject by knowing the total study time available.

In the given exercise, with the constraint \(x + y = 1\), the substitution method rewrites the function \(f(x,y)\) as \(f(x, 1-x)\), thus turning the problem into a single-variable issue. The key is to replace \(y\) with \(1-x\), making the problem much more manageable. Substitution is especially handy when dealing with two variables, as it allows standard calculus techniques, like differentiation, to be applied directly.

Critical Points

In calculus, critical points are where a function's derivative equals zero or is undefined, marking potential maximums, minimums, or inflection points. Think of a drive through hilly roads: the top of a hill or the bottom of a valley—those are like critical points, the high and low of the function's landscape.

To identify the critical points for the function \(f(x, 1-x) = x - x^2\), we calculate the first derivative and set it equal to zero. From the exercise, setting \(1 - 2x = 0\), we deduce that \(x = 1/2\) is the critical point to consider. In multidimensional settings, critical points are determined by finding where all partial derivatives vanish simultaneously or where gradients become zero vectors, indicating no immediate increase or decrease in any direction.

First Derivative Test

The first derivative test helps decide whether a critical point is a local maximum, minimum, or neither. Remember the hilly road? The first derivative test is akin to feeling the slope with your feet to tell if you're at the top or bottom of a hill.

To apply the first derivative test, we calculate the derivative and then examine its sign to the left and right of the critical point. A sign change from positive to negative as you pass through the critical point suggests a local maximum, while a change from negative to positive indicates a local minimum. In the provided exercise, the signs around \(x = 1/2\) indicate a maximum.

Second Derivative Test

The second derivative test offers another way to determine the nature of critical points. It essentially tells us about the curvature of the function near a critical point, akin to identifying if you're on the upwards or downwards curving part of a hill.

If the second derivative at a critical point is positive, the function has a concave up shape there, indicating a local minimum. Conversely, a negative second derivative suggests a concave down shape and thus a local maximum. In our exercise, the second derivative is -2 at the critical point \(x = 1/2\), confirming it as a maximum. However, this test isn't conclusive when the second derivative equals zero, as it doesn't provide information about curvature in such cases.