Stationary Values And Points Investigation In Multivariable Calculus
In this section, we delve into the process of finding the stationary values of the given function z = x² + xy + y² + 5x - 5y + 3. Stationary points, also known as critical points, are points where the function's rate of change is zero. These points are crucial in optimization problems, as they can represent local maxima, local minima, or saddle points. To find these stationary values, we need to find the points where the partial derivatives of the function with respect to x and y are simultaneously equal to zero. This involves a systematic approach using calculus and algebraic manipulation. We will explore the steps involved in finding these points and how to classify them using the second derivative test. Understanding stationary values is fundamental in various fields, including physics, engineering, and economics, where optimization is a key aspect of problem-solving. By mastering the techniques to identify and classify stationary points, one can gain a deeper understanding of the behavior of functions and their applications in real-world scenarios. The process begins with calculating the partial derivatives and setting them to zero, leading to a system of equations. Solving this system will give us the coordinates of the stationary points. Further analysis using the second derivative test will then help us determine the nature of these points, whether they represent a maximum, minimum, or saddle point. This comprehensive analysis provides valuable insights into the function's behavior and its potential applications.
First, we need to find the partial derivatives of z with respect to x and y. The partial derivative of z with respect to x, denoted as ∂z/∂x, is found by differentiating z with respect to x while treating y as a constant. Similarly, the partial derivative of z with respect to y, denoted as ∂z/∂y, is found by differentiating z with respect to y while treating x as a constant. These partial derivatives represent the rate of change of the function in the x and y directions, respectively. Setting these derivatives to zero allows us to find the points where the function's rate of change is zero in both directions, indicating a potential stationary point. The calculations are as follows:
∂z/∂x = 2x + y + 5 ∂z/∂y = x + 2y - 5
To find the stationary points, we set both partial derivatives equal to zero:
2x + y + 5 = 0 x + 2y - 5 = 0
This system of linear equations can be solved using various methods, such as substitution or elimination. Let's use the method of substitution. From the first equation, we can express y in terms of x: y = -2x - 5
Substitute this expression for y into the second equation: x + 2(-2x - 5) - 5 = 0 x - 4x - 10 - 5 = 0 -3x - 15 = 0 -3x = 15 x = -5
Now, substitute x = -5 back into the equation for y: y = -2(-5) - 5 y = 10 - 5 y = 5
Thus, the stationary point is (-5, 5). To classify this stationary point, we need to compute the second partial derivatives and use the second derivative test. The second partial derivatives are:
∂²z/∂x² = 2 ∂²z/∂y² = 2 ∂²z/∂x∂y = 1
The discriminant, D, is given by: D = (∂²z/∂x²)(∂²z/∂y²) - (∂²z/∂x∂y)² D = (2)(2) - (1)² D = 4 - 1 D = 3
Since D > 0 and ∂²z/∂x² > 0, the stationary point (-5, 5) is a local minimum. The value of z at this point is:
z = (-5)² + (-5)(5) + (5)² + 5(-5) - 5(5) + 3 z = 25 - 25 + 25 - 25 - 25 + 3 z = -22
Therefore, the function z = x² + xy + y² + 5x - 5y + 3 has a local minimum at the point (-5, 5) with a value of -22.
In this section, we address the problem of finding the stationary point of the function F = x² + y² subject to the constraint x² + y² + 2x - 2y + 1 = 0. This is a classic example of an optimization problem with constraints, which can be effectively solved using the method of Lagrange multipliers. This method allows us to find the extreme values of a function subject to one or more constraints by introducing a new variable, the Lagrange multiplier, to form a Lagrangian function. The stationary points of the Lagrangian function correspond to the stationary points of the original function subject to the constraint. Understanding Lagrange multipliers is essential in various fields, including economics, engineering, and physics, where optimization under constraints is a common problem. We will walk through the steps of setting up the Lagrangian function, finding its partial derivatives, and solving the resulting system of equations to determine the stationary point. This process involves careful algebraic manipulation and a solid understanding of calculus principles. By mastering this technique, one can tackle a wide range of optimization problems where constraints play a crucial role in the solution.
To begin, we need to form the Lagrangian function. This function combines the original function F and the constraint equation using a Lagrange multiplier, typically denoted by λ (lambda). The Lagrangian function, L, is defined as:
L(x, y, λ) = F(x, y) - λ * g(x, y)
where F(x, y) = x² + y² is the function to be optimized, and g(x, y) = x² + y² + 2x - 2y + 1 = 0 is the constraint equation. Thus, the Lagrangian function is:
L(x, y, λ) = x² + y² - λ(x² + y² + 2x - 2y + 1)
Next, we need to find the partial derivatives of L with respect to x, y, and λ and set them equal to zero. These partial derivatives represent the rates of change of the Lagrangian function in each direction. Setting them to zero allows us to find the critical points of the Lagrangian function, which correspond to the stationary points of the original function subject to the constraint. The partial derivatives are:
∂L/∂x = 2x - λ(2x + 2) ∂L/∂y = 2y - λ(2y - 2) ∂L/∂λ = -(x² + y² + 2x - 2y + 1)
Setting these partial derivatives to zero gives us the following system of equations:
2x - λ(2x + 2) = 0 (1) 2y - λ(2y - 2) = 0 (2) x² + y² + 2x - 2y + 1 = 0 (3)
From equation (1), we have:
2x = λ(2x + 2) x = λ(x + 1) (4)
From equation (2), we have:
2y = λ(2y - 2) y = λ(y - 1) (5)
If λ = 0, then from (4) x = 0 and from (5) y = 0. Substituting x = 0 and y = 0 into equation (3):
0² + 0² + 2(0) - 2(0) + 1 = 0 1 = 0
This is a contradiction, so λ ≠ 0.
From equations (4) and (5), we have:
λ = x / (x + 1) λ = y / (y - 1)
Equating the expressions for λ:
x / (x + 1) = y / (y - 1) x(y - 1) = y(x + 1) xy - x = xy + y -x = y
Substitute y = -x into equation (3):
x² + (-x)² + 2x - 2(-x) + 1 = 0 x² + x² + 2x + 2x + 1 = 0 2x² + 4x + 1 = 0
Using the quadratic formula to solve for x:
x = (-b ± √(b² - 4ac)) / (2a) x = (-4 ± √(4² - 4(2)(1))) / (2(2)) x = (-4 ± √(16 - 8)) / 4 x = (-4 ± √8) / 4 x = (-4 ± 2√2) / 4 x = -1 ± (√2) / 2
So, we have two possible values for x:
x₁ = -1 + (√2) / 2 x₂ = -1 - (√2) / 2
Since y = -x, the corresponding values for y are:
y₁ = 1 - (√2) / 2 y₂ = 1 + (√2) / 2
Thus, the stationary points are:
(-1 + (√2) / 2, 1 - (√2) / 2) (-1 - (√2) / 2, 1 + (√2) / 2)
To determine which point minimizes F = x² + y², we can evaluate F at each point:
F(x₁, y₁) = (-1 + (√2) / 2)² + (1 - (√2) / 2)² = 2(1 - (√2) + 1/2) = 2(3/2 - √2) = 3 - 2√2 F(x₂, y₂) = (-1 - (√2) / 2)² + (1 + (√2) / 2)² = 2(1 + (√2) + 1/2) = 2(3/2 + √2) = 3 + 2√2
Since 3 - 2√2 < 3 + 2√2, the stationary point that minimizes F is (-1 + (√2) / 2, 1 - (√2) / 2).
Therefore, the stationary point of the function F = x² + y² subject to the constraint x² + y² + 2x - 2y + 1 = 0 is (-1 + (√2) / 2, 1 - (√2) / 2).
