Question

Determine the stationary points of \(f(x, y)=\) \(2 x^{2}+3 y^{2}+5 x+12 y+19\)

Short answer

Answer: The stationary point of the given function is at \((-\frac{5}{4}, -2)\).

Step-by-step solution

Calculate the partial derivatives with respect to x and y

To find the partial derivative of f with respect to x, differentiate the function with respect to x while treating y as a constant: \(\frac{\partial f}{\partial x} = 4x + 5\). Similarly, to find the partial derivative of f with respect to y, differentiate the function with respect to y while treating x as a constant: \(\frac{\partial f}{\partial y} = 6y + 12\).

Set the partial derivatives equal to zero

To locate the stationary points, we set both partial derivatives equal to zero: \(4x + 5 = 0\) and \(6y + 12 = 0\).

Solve the system of equations

To solve the equations, first solve each one for the variable involved: For the equation regarding x: \(4x + 5 = 0\), subtract 5 from both sides and then divide by 4 \(x = -\frac{5}{4}\). For the equation regarding y: \(6y + 12 = 0\), subtract 12 from both sides and then divide by 6 \(y = -2\). The stationary point of \(f(x, y) = 2x^2 + 3y^2 + 5x + 12y + 19\) is at the point \((-\frac{5}{4}, -2)\).

Key concepts

Partial Derivatives

Partial derivatives are essential in multivariable calculus when dealing with functions of more than one variable. If you have a function like \( f(x, y) \), which depends on two independent variables, you can understand how the function changes with respect to each variable separately by using partial derivatives.

• The partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), measures how \( f \) changes as \( x \) changes, while keeping \( y \) constant.
• Similarly, \( \frac{\partial f}{\partial y} \) tells you how \( f \) changes with \( y \), holding \( x \) steady.

To calculate partial derivatives, you treat the variable you are not differentiating with respect to as a constant. For example, if you want \( \frac{\partial f}{\partial x} \) for \( f(x, y) = 2x^2 + 3y^2 + 5x + 12y + 19 \), you derive it as if it were just \( 2x^2 + 5x \) and forget about the \( y \) terms initially.
Once both partial derivatives are computed, setting them to zero helps locate stationary points, where the slope is neither increasing nor decreasing, indicating potential minima, maxima, or saddle points.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to functions of multiple variables, like calculating gradients, multivariable limits, and Taylor series expansions. In the context of finding stationary points, multivariable calculus techniques become crucial. A stationary point is where any change in any of the input variables does not change the function value—these points are neither rising nor falling.
When you find a stationary point for \( f(x, y) \), you use both partial derivatives. If \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \), you are identifying a place where the function \( f \) does not 'tilt' in the \( x \) or \( y \) direction.

• This concept helps in identifying all types of extremas (local minima, maxima) or testing for saddle points via second derivative tests called the Hessian determinant.
• Setting up these tests requires understanding the interactions between different variables and applying calculus to functions defined on more dimensional spaces like \( \mathbb{R}^2 \).

Thus, multivariable calculus provides the tools necessary to extend the powerful concepts of calculus into dimensions higher than one.

System of Equations

A system of equations involves simultaneously dealing with multiple equations to find the solutions for variables present in those equations. When it comes to stationary points, systems of equations arise naturally as part of the solution.
In our case, after calculating the partial derivatives for \( f(x, y) \), we obtain a system:

• \( 4x + 5 = 0 \)
• \( 6y + 12 = 0 \)

To find \( x \) and \( y \), you solve the equations simultaneously:

• For \( 4x + 5 = 0 \): move 5 to the opposite side \( 4x = -5 \), and dividing by 4 gives \( x = -\frac{5}{4} \).
• For \( 6y + 12 = 0 \): subtracting 12 gives \( 6y = -12 \), and dividing by 6 results in \( y = -2 \).

The solutions \( x = -\frac{5}{4} \) and \( y = -2 \) indicate where both derivatives are zero simultaneously, marking the stationary point. Solving such systems is often straightforward with linear equations, but can be more complex with higher degree polynomials or nonlinear terms, requiring advanced techniques like substitution or elimination methods.