Chapter 6: Problem 24
Find \(f_{x}, f_{y},\) and \(f_{\lambda}\) $$f(x, y, \lambda)=9 x y-\lambda(3 x-y+7)$$
Short Answer
Expert verified
\(f_x = 9y - 3\lambda\), \(f_y = 9x + \lambda\), \(f_{\lambda} = -(3x - y + 7)\)
Step by step solution
01
Identify Partial Derivative with respect to x
To find the partial derivative of the function with respect to x, we treat y and \lambda as constants. The original function is \(f(x, y, \lambda) = 9xy - \lambda(3x - y + 7)\). Using the rules of differentiation, we find the derivative with respect to x: \[f_x = \frac{\partial (9xy - \lambda(3x - y + 7))}{\partial x} = 9y - \lambda(3) = 9y - 3\lambda\].
02
Identify Partial Derivative with respect to y
Next, we find the partial derivative of the function with respect to y by treating x and \lambda as constants. Thus, \[f_y = \frac{\partial (9xy - \lambda(3x - y + 7))}{\partial y} = 9x - (-\lambda) = 9x + \lambda\].
03
Identify Partial Derivative with respect to \lambda
Finally, we determine the partial derivative with respect to \lambda, by treating x and y as constants. The derivative becomes: \[f_{\lambda} = \frac{\partial (9xy - \lambda(3x - y + 7))}{\partial \lambda} = -(3x - y + 7)\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation
Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes at any given point. It's like measuring how fast something is moving at a specific moment in time. When we perform differentiation, we calculate the derivative of a function.
Let's illustrate this using the partial derivatives from our exercise:
Let's illustrate this using the partial derivatives from our exercise:
- Partial derivative with respect to x: This tells us how the function changes as x changes, while keeping y and \lambda constant. For the function provided: \(f_x = \frac{\text d}{\text dx} (f(x, y, \lambda)) = 9y - 3\lambda \). This is derived by focusing solely on the terms involving x.
- Partial derivative with respect to y: This tells us how the function changes as y changes, while x and \lambda are constant. Using the provided function: \(f_y = \frac{\text d}{\text dy} (f(x, y, \lambda)) = 9x + \lambda \). This is obtained by focusing on the terms involving y.
- Partial derivative with respect to lambda (\lambda): This measures the sensitivity of the function with respect to changes in \lambda, keeping both x and y constant. For the function given: \(f_\lambda = \frac{\text d}{\text d\lambda} (f(x, y, \lambda)) = -(3x - y + 7) \).
Multivariable Calculus
Multivariable calculus extends single-variable calculus to functions of multiple variables, such as the one given in our exercise. Unlike single-variable functions which only depend on one input, multivariable functions depend on two or more inputs.
Consider the function from our exercise: \(f(x, y, \lambda) = 9xy - \lambda(3x - y + 7) \). This is a function of three variables: x, y, and \lambda. Here are some key aspects of multivariable calculus:
Consider the function from our exercise: \(f(x, y, \lambda) = 9xy - \lambda(3x - y + 7) \). This is a function of three variables: x, y, and \lambda. Here are some key aspects of multivariable calculus:
- Gradient: The gradient of a multivariable function is a vector that points in the direction of the greatest rate of increase of the function. It's composed of the partial derivatives with respect to each variable. For our function: \( abla f = ( f_x, f_y, f_\lambda ) = (9y - 3\lambda, 9x + \lambda, -(3x-y+7)) \).
- Level Sets: These are sets of points where the function takes on a constant value. For two variables, these are often curves (like contours on a map), and for three variables, they can be surfaces in space.
Lagrange Multipliers
Lagrange multipliers are a strategy used in optimization problems to find the local maxima and minima of a function subject to equality constraints. This technique is ideal when dealing with multivariable functions like the one in our exercise.
The basic idea behind Lagrange multipliers is to convert a constrained problem into an unconstrained one by introducing new variables called multipliers. Here's a breakdown:
The basic idea behind Lagrange multipliers is to convert a constrained problem into an unconstrained one by introducing new variables called multipliers. Here's a breakdown:
- Objective Function: The function you want to maximize or minimize, denoted as L(x, y, \lambda).
- Constraint: The condition that must be satisfied, typically given as g(x, y) = 0.
- Lagrangian Function: Combines the objective function and the constraint using a multiplier \lambda: \L(x, y, \lambda) = f(x, y) - \lambda g(x, y).