Problem 10 - Calculus an Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=x^{2}+6 x y+10 y^{2}-4 y+4 $$

Step-by-Step Solution

Answer

The function \(f(x, y) = x^{2} + 6xy + 10y^{2} - 4y + 4\) has a relative minimum at the point \((x, y) = \left(-\frac{1}{2}, \frac{1}{5}\right)\). It has no saddle points.

Step 1: Find the First-Order Partial Derivatives

The first order partial derivatives of \(f(x, y)\) are given by: \(f_{x} = \frac{\partial f}{\partial x} = 2x + 6y\), \(f_{y} = \frac{\partial f}{\partial y} = 6x + 20y - 4\).

Step 2: Find the Critical Points

Set the first order partial derivatives to zero to find the potential critical points by solving the following system of equations: \(2x + 6y = 0\), \(6x + 20y - 4 = 0\). Solving this system, one finds the critical point to be \((x, y) = \left(-\frac{1}{2}, \frac{1}{5}\right)\).

Step 3: Apply the Second Partial Derivative Test

Calculate the second order partial derivatives: \(f_{xx} = \frac{\partial^{2} f}{\partial x^{2}} = 2\), \(f_{xy} = \frac{\partial^{2} f}{\partial x\partial y} = 6\), and \(f_{yy} = \frac{\partial^{2} f}{\partial y^{2}} = 20\). Then calculate the determinant of the Hessian matrix \(D = f_{xx}f_{yy} - f_{xy}^{2} = 2*20 - 6^{2} = 4 > 0\). Also, \(f_{xx} > 0\), therefore the critical point \((x, y) = \left(-\frac{1}{2}, \frac{1}{5}\right)\) is a relative minimum.

Key Concepts

Critical Points in Multivariable Calculus

In multivariable calculus, critical points are essential in understanding the highs and lows of a function’s landscape. To picture this, imagine a hiker seeking a peak or valley in a range of hills; critical points are the coordinates where she might find what she's looking for. These critical points are locations on surface plots where the function's output doesn't increase or decrease when you make infinitesimally small moves in any direction.

How do we find them? By harnessing the power of first-order partial derivatives. At critical points, all these derivatives simultaneously vanish. This means we set each partial derivative equal to zero and solve for the variables involved. The resulting solutions tell us the coordinates of our critical points - the candidates for peaks, valleys, or more complex topographical features.

First-Order Partial Derivatives

When faced with a multivariable function like our example, you can think of first-order partial derivatives as the function's heartbeats in each direction of space. For a function of two variables, these derivatives gauge how sensitive the function is to slight changes in either direction – east-west or north-south, so to speak.

In technical terms, a first-order partial derivative represents the slope of the tangent line to a curve you'd get if you slice through the function's graph while keeping the other variable fixed. Solving \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 0 \) together is like plotting the function’s anatomy – it reveals where the heartbeats soften to a silence, unveiling possible critical points.

Hessian Matrix in Calculus

After locating critical points, we must ask: What nature do they possess? Are they peaks, valleys, or mere flat spots? Enter the Hessian matrix, a tool akin to an MRI scan for the function's topology. This matrix is assembled from second-order partial derivatives of the function, offering a deeper look into its curvature characteristics at any given point.

The determinant of the Hessian, observed at a critical point, decides the point's fate. If the determinant, symbolized as \(D\), is positive, and the second-order derivative with respect to \(x\),\(f_{xx}\), is positive, it points to a valley – scientifically called a relative minimum. Conversely, if \(f_{xx}\) is negative, the critical point takes the form of a peak – a relative maximum. Should the determinant tread into negative territories, the critical point becomes a 'saddle point', an interesting terrain where the function is cupped upwards in one direction and downwards in another.

Relative Extrema

Peering into the realms of relative extrema is like seeking out the most scenic spots along the hiker's journey. These are the points where locally, there are no higher peaks immediately nearby (relative maxima) or lower valleys (relative minima). They are essential in optimization problems where finding the best solution is akin to finding the highest or lowest point in a landscape.

Diagnosing Relative Extrema

Armed with the second partial derivative test, we diagnose relative extrema by analyzing the temperature of the Hessian matrix. Steady and positive? You're likely sitting in a cozy valley. A descent into negatives meanwhile could signal a majestic peak, depending on the sign of \(f_{xx}\). It's important to remember that these assessments are localized – there might be even higher peaks or lower valleys elsewhere in the function's domain. Yet, understanding where these extrema occur is a powerful guide to navigating the function's terrain.