Problem 11 - Calculus an Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=3 x^{2}+2 y^{2}-12 x-4 y+7 $$

Step-by-Step Solution

Answer

The function \(f(x, y)=3 x^{2}+2 y^{2}-12 x-4 y+7\) has a relative minimum at the point \((2, 1)\).

Step 1: Find the Partial Derivatives

The first step is finding the first order partial derivatives of the function \(f(x,y)\). The partial derivatives are given by: \(\frac{\partial f}{\partial x} = 6x - 12\) and \(\frac{\partial f}{\partial y} = 4y - 4.\)

Step 2: Solve Equations to Find Critical Points

Next, to find the critical points, one must solve the system of equations where the partial derivatives are both equal to zero. The system of equations is: \(\frac{\partial f}{\partial x} = 0\) and \(\frac{\partial f}{\partial y} = 0\). Solving this gives us the critical points, which are \((x, y) = (2, 1)\).

Step 3: Calculate the Second Order Partial Derivatives

Calculate the second order partial derivatives next. They are \(\frac{\partial^{2} f}{\partial x^{2}}=6\), \(\frac{\partial^{2} f}{\partial y^{2}}=4\), and \(\frac{\partial^{2} f}{\partial x \partial y}=0\).

Step 4: Analyze the Critical Points

To test if these critical points are relative extrema or saddle points, construct the Hessian matrix from the second order partial derivatives and calculate its determinant. The determinant \((\frac{\partial^{2} f}{\partial x^{2}} * \frac{\partial^{2} f}{\partial y^{2}}) - (\frac{\partial^{2} f}{\partial x \partial y})^{2} = (6*4) - (0) = 24\) is positive, implying that the critical point is a relative extremum. As \(\frac{\partial^{2} f}{\partial x^{2}}=6\) and \(\frac{\partial^{2} f}{\partial y^{2}}=4\) are both positive, therefore it is a relative minimum.

Key Concepts

Partial Derivatives

Understanding partial derivatives is crucial when studying functions of multiple variables, like the given function \(f(x, y)=3 x^{2}+2 y^{2}-12 x-4 y+7\). In essence, a partial derivative represents how a function changes as only one of the variables changes, holding the others fixed. Imagine you're on a hillside; looking horizontally across the hill, you see how steep the hill is in the east-west direction—that's akin to a partial derivative with respect to \(x\). Looking in the north-south direction gives you another slope, which is analogous to the partial derivative with respect to \(y\).

For the function \(f(x, y)\), we calculate the partial derivatives with respect to \(x\) and \(y\). Respectively, these are given by the formulas \(\frac{\partial f}{\partial x} = 6x - 12\) and \(\frac{\partial f}{\partial y} = 4y - 4\). These formulas show us the 'slope' of the function in each direction at any point \((x, y)\).

To put it simply, the steeper the slope, the greater the rate of increase or decrease of the function as you move in the direction of that variable. This is why finding partial derivatives is the first step and a fundamental part of identifying both relative extrema and saddle points.

Critical Points

The notion of critical points is another step towards understanding extrema and saddle points of a function. A critical point occurs when all the first order partial derivatives of the function are equal to zero. This indicates a potential 'flat' spot, such as the top or bottom of a hill, or a saddle point, which could represent a mountain pass.

For the given function \(f(x, y)\), we find these points by setting the partial derivatives to zero: \(\frac{\partial f}{\partial x} = 0\) and \(\frac{\partial f}{\partial y} = 0\). This system of equations, when solved, yields the critical points: in our case, \((x, y) = (2, 1)\). At this point, the 'slopes' in both directions are flat, suggesting a relative extremum or saddle point at that location. Identifying and testing these points are paramount to understanding the nature of the function's graph and discerning where it has its peaks, valleys, and passes.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a multivariable function. It is pivotal for determining the concavity at a critical point and whether that point is a relative minimum, relative maximum, or a saddle point. Much like how geomorphologists study the curves of landscapes to identify peaks and valleys, mathematicians use the Hessian to study the curvature of a function.

In this scenario, the Hessian matrix for the function \(f(x, y)\) would be constructed using the second-order partial derivatives \(\frac{\partial^{2} f}{\partial x^{2}}=6\), \(\frac{\partial^{2} f}{\partial y^{2}}=4\), and \(\frac{\partial^{2} f}{\partial x \partial y}=0\). The determinant of this Hessian matrix at the critical point \((2, 1)\), given by \((6*4) - (0)^2 = 24\), is positive, which is characteristic of a relative extremum. Furthermore, since both second-order partial derivatives are positive, the function is concave up at that point, much like a bowl, indicating a relative minimum. In layman's terms, at this particular spot on the function's 'terrain', all paths lead uphill, confirming it as a local low point.