Problem 23 - Calculus and its applications

Question

Find \(f_{x}, f_{y},\) and \(f_{\lambda}\). $$f(x, y, \lambda)=5 x y-\lambda(2 x+y-8)$$

Step-by-Step Solution

Answer

\( f_{x} = 5y - 2\lambda \), \( f_{y} = 5x - \lambda \), \( f_{\lambda} = -(2x + y - 8) \)

Step 1: Differentiate with respect to x

First, we want to find the partial derivative of the function with respect to \(x\). Apply the rules of differentiation to the function \(f(x, y, \lambda) = 5xy - \lambda(2x + y - 8)\).\[ f_{x} = \frac{\partial}{\partial x}(5xy - \lambda(2x + y - 8)) \]Differentiate each term with respect to \(x\):- \( \frac{\partial}{\partial x}(5xy) = 5y \)- \( \frac{\partial}{\partial x}(-\lambda(2x + y - 8)) = -\lambda(2) = -2\lambda \)So,\[ f_{x} = 5y - 2\lambda \]

Step 2: Differentiate with respect to y

Next, find the partial derivative of the function with respect to \(y\). Differentiate as follows:\[ f_{y} = \frac{\partial}{\partial y}(5xy - \lambda(2x + y - 8)) \]Perform the differentiation:- \( \frac{\partial}{\partial y}(5xy) = 5x \)- \( \frac{\partial}{\partial y}(-\lambda(2x + y - 8)) = -\lambda \)Thus, the partial derivative with respect to \(y\) is:\[ f_{y} = 5x - \lambda \]

Step 3: Differentiate with respect to λ

Finally, differentiate the function with respect to \(\lambda\):\[ f_{\lambda} = \frac{\partial}{\partial \lambda}(5xy - \lambda(2x + y - 8)) \]This gives:- The derivative of \(5xy\) with respect to \(\lambda\) is 0, since \(5xy\) does not contain \(\lambda\).- \( \frac{\partial}{\partial \lambda}(-\lambda(2x + y - 8)) = -(2x + y - 8) \)Thus, the partial derivative with respect to \(\lambda\) is:\[ f_{\lambda} = -(2x + y - 8) \]

Key Concepts

Multivariable Calculus

In calculus, when dealing with functions of more than one variable, like in our original exercise, we enter the realm of multivariable calculus. This branch of mathematics extends the principles of differential and integral calculus to functions of multiple variables. Rather than a single input and output, you deal with several inputs, and sometimes, multiple outputs as well. The function in our exercise is a multivariable function defined as follows: \[ f(x, y, \lambda) = 5xy - \lambda(2x + y - 8) \]Here, the function is dependent on three variables: \(x\), \(y\), and the Lagrange multiplier \(\lambda\). Such functions are common in scenarios where you model relationships involving diverse elements, such as economic models, physical systems, and optimization problems. Understanding multivariable functions allows us to explore how small changes in one variable can affect the function's output, or how these variables interact with each other.

Differentiation Rules

Differentiation is the process of finding a derivative, which is a measure of how a function changes as its input changes. In our exercise, we find partial derivatives, which represent the rate of change of the multivariable function with respect to one of its variables while keeping the other variables constant.To compute partial derivatives, some core differentiation rules are applied:

• Constant Multiplication Rule: To differentiate a term like \(c \times u\), where \(c\) is a constant, simply take the derivative of \(u\) and multiply by the constant \(c\).
• Power Rule: For terms of the form \(x^n\), differentiate as \(nx^{n-1}\).
• Sum/Difference Rule: The derivative of a sum or difference is the sum or difference of their derivatives.

In this exercise:- For \(f_{x}\), we find derivatives like \(\frac{\partial}{\partial x}(5xy) = 5y\).- We also apply rules like \(\frac{\partial}{\partial x}(-\lambda(2x + y - 8)) = -2\lambda\), using the constant multiplier method because \(\lambda\) is treated as a constant when differentiating with respect to \(x\).By understanding these rules, finding partial derivatives becomes a straightforward task, allowing us to derive insights about the function's behavior. This skill is crucial in a variety of scientific and engineering fields.

Lagrange Multipliers

Lagrange multipliers are a powerful mathematical method used for finding the local maxima and minima of a function subject to equality constraints. This technique is particularly useful in optimization problems where you need to maximize or minimize a function given some conditions or restrictions.In the context of the exercise, we include the term \(\lambda(2x + y - 8)\) in the function, where \(\lambda\) is the Lagrange multiplier. This inclusion allows us to embed the constraint directly into the function. The idea behind using Lagrange multipliers is to convert a constrained optimization problem into an unconstrained one by incorporating the constraints into the function itself. The steps generally involve:

• Defining the Lagrangian function, which adds the product of the Lagrange multiplier \(\lambda\) and the constraint to the original function.
• Finding partial derivatives with respect to each variable and the Lagrange multiplier.
• Setting these derivatives equal to zero and solving the system of equations to find the optimal points.

By understanding and applying the Lagrange multiplier method, we can solve complex optimization problems that appear in engineering, finance, and economics effectively. This method is an invaluable tool for analyzing functions that need to consider constraints.