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Chapter 2: Problem 6

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 Langrangian multiplier method.

Short Answer

Expert verified
- The maximum value of the function is \(\frac{1}{4}\) found using both the substitution and Lagrangian multiplier methods.

Step by step solution

01

Substitution Method

Eliminate one variable using the constraint \(x + y = 1\). We can express \(y\) in terms of \(x\): \(y = 1 - x\). Now substitute this into the function \(f(x, y)\): \(f(x, 1 - x) = x(1 - x)\).
02

Find the critical points

To find the maximum of the function within the constraint, we need to find the critical points, meaning the points where the derivative of the function is zero or undefined. Find the derivative of the function \(f\) with respect to \(x\): \(f'(x) = \frac{d}{dx} (x(1 - x)) = 1 - 2x\).
03

Find the maximum value

Now, we need to find the maximum value of \(f(x)\). To do this, set the derivative equal to 0 and solve for x: \(1 - 2x = 0 \Rightarrow x = \frac{1}{2}\). Now, recall that we have the constraint \(x + y = 1\). Substituting \(x = \frac{1}{2}\), we obtain \(y = \frac{1}{2}\). Now, we can find the maximum value of the function: \(f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\).
04

Lagrangian Multiplier Method

Now, we will use the Lagrangian multiplier method to find the maximum value of the function \(f(x, y) = xy\) with the constraint \(x + y = 1\). First, we must define our Lagrangian function. The Lagrangian function, denoted by \(L\), is defined as follows: \(L(x, y, \lambda) = f(x, y) - \lambda(x + y - 1)\), where \(\lambda\) is the Lagrangian multiplier.
05

Computing the gradient

To find the critical points, we need to find the gradient of the Lagrangian function and set each component equal to zero: $\nabla L (x, y, \lambda) = \begin{cases} \frac{\partial L}{\partial x} = y - \lambda = 0\\ \frac{\partial L}{\partial y} = x - \lambda = 0 \\ \frac{\partial L}{\partial \lambda} = x + y - 1 = 0 \end{cases}$.
06

Solving the system of equations

Now, we need to solve the system of equations: \begin{cases} y - \lambda = 0\\ x - \lambda = 0 \\ x + y - 1 = 0 \end{cases}. Statement 1 implies \(\lambda = y\), and statement 2 implies \(\lambda = x\). Therefore, \(x = y\). From statement 3, we obtain \(2x - 1 = 0\) or \(x = \frac{1}{2}\). Thus, \(y = \frac{1}{2}\) as well.
07

Find the maximum value

With the solutions \(x = \frac{1}{2}\) and \(y = \frac{1}{2}\), we can find the maximum value of the function \(f(x, y)\): \(f\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}\). Both methods, substitution and Lagrangian multiplier, found the same maximum value for the function, \(\frac{1}{4}\).

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Most popular questions from this chapter

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