Question

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

Short answer

Answer: The maximum value of the function is \(\frac{1}{4}\) at the point \((\frac{1}{2},\frac{1}{2})\).

Step-by-step solution

Method 1: Substitution

Since we know that \(x+y=1\), we can easily solve for one variable in terms of the other. Let's solve for \(y\): $$y = 1-x$$ Now, we can substitute this expression for \(y\) in the function \(f(x, y)\): $$f(x, 1-x) = x(1-x)$$ To find the maximum value, we should find the critical points by taking the derivative of the function with respect to \(x\) and set it to zero: $$\frac{d}{dx}(x(1-x)) = (1-x)-x = 1-2x$$ Now, let's find the critical points: $$1-2x=0 \Rightarrow x=\frac{1}{2}$$ Substitute back into \(y\) expression: $$y=1-x \Rightarrow y=1-\frac{1}{2}=\frac{1}{2}$$ The maximum value for \(f(x,y)\) is at the point \((\frac{1}{2},\frac{1}{2})\), so: $$f_{max}(x, y) = f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2} * \frac{1}{2} = \frac{1}{4}$$

Method 2: Lagrangian Multipliers

To solve the problem by using the Lagrangian multiplier method, we first set up the Lagrangian function with a multiplier \(\lambda\), let's denote \(g(x,y)=x+y\): $$\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda(g(x,y)-1) = xy - \lambda(x+y-1)$$ Now, in order to find the maximum value, we have to find where the gradient of the Lagrangian is equal to zero: $$ \nabla \mathcal{L}(x, y, \lambda) = \begin{bmatrix} \frac{\partial \mathcal{L}}{\partial x}\\ \frac{\partial \mathcal{L}}{\partial y}\\ \frac{\partial \mathcal{L}}{\partial \lambda} \end{bmatrix} = \begin{bmatrix} y - \lambda \\ x - \lambda \\ x + y - 1 \end{bmatrix} = 0$$ Solve these equations simultaneously for \(x\), \(y\), and \(\lambda\): $$\begin{cases} y - \lambda = 0 \\ x - \lambda = 0 \\ x + y - 1 = 0 \end{cases}$$ From the first two equations, we can conclude that \(x=y\), and the third equation becomes: $$x+x-1=0 \Rightarrow x=\frac{1}{2}$$ Substitute back into the first equation: $$y - \lambda = 0 \Rightarrow y=\lambda=\frac{1}{2}$$ The maximum value for \(f(x,y)\) is obtained at the point \((\frac{1}{2},\frac{1}{2})\), so: $$f_{max}(x, y) = f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2} * \frac{1}{2} = \frac{1}{4}$$ Both methods yield the same maximum value of \(\frac{1}{4}\) at the point \((\frac{1}{2},\frac{1}{2})\).

Key concepts

Lagrangian multiplier method

The Lagrangian multiplier method is a powerful tool used in constrained optimization; it's quite the life-saver for problems that involve maximizing or minimizing a function subject to constraints. This method works by transforming a constrained problem into an unconstrained one.
Here's a simple-to-understand breakdown of how it works:

• We incorporate the constraint into the function we want to optimize by introducing a new variable called the Lagrange multiplier, denoted by \( \lambda \).
• We set up a new function, the Lagrangian, which includes the original function and the constraint multiplied by \( \lambda \).
• In the case of this exercise, the Lagrangian is \( \mathcal{L}(x, y, \lambda) = xy - \lambda(x+y-1) \).
• To find the maximum or minimum values, we find the points where the gradient of the Lagrangian equals zero, which involves solving a set of equations.

In our example, these equations help us find the optimal values of \( x \) and \( y \) that satisfy the constraint \( x + y = 1 \), giving us \( x = y = \frac{1}{2} \). Then we can substitute these values back into the original function to find the maximum value.

Substitution method

The substitution method is often the student's first encounter with solving constrained optimization problems.
It's straightforward and involves replacing one of the variables in our function with another expression derived from the constraint.
In the problem presented:

• We initially know for our constraint that \( x + y = 1 \). So, we can express \( y \) in terms of \( x \), as \( y = 1 - x \).
• This substitution is used in our function \( f(x, y) = xy \), which then becomes \( x(1-x) \).
• After substitution, the problem reduces to finding the maximum of this single-variable function.

To find that maximum value, we'll take the derivative of the resultant expression, set it to zero, and solve for \( x \).
In this case, \( \frac{d}{dx}(x - x^2) = 1 - 2x \), giving us \( x = \frac{1}{2} \) when set to zero.
From our constraint \( y = 1 - x \), \( y \) also equals \( \frac{1}{2} \).
This gives us the maximum value of \( \frac{1}{4} \) at \( x = y = \frac{1}{2} \).

Maximum value

Finding the maximum value of a function subject to constraints is a common task, especially in fields such as economics and engineering.
In this exercise, the goal was to determine the highest possible value of the function \( f(x, y) = xy \) given the constraint \( x+y=1 \).
Both the Lagrangian multiplier method and the substitution method aim to identify the critical points where the function could attain a maximum or minimum under the constraints.

• Both methods ultimately identified the point \((\frac{1}{2}, \frac{1}{2})\) as the one where the maximum value occurs.
• By substituting \( x = \frac{1}{2} \) and \( y = \frac{1}{2} \) into the original function \( xy \), we discern the function peaks at \( \frac{1}{4} \).

These two methods provide a confirmation of each other's work, affirming that the function's maximal value under the specified constraint is indeed \( \frac{1}{4} \).
This exploration helps solidify the understanding of solving constrained optimization problems effectively.