Problem 315 - Calculus Volume 3

Question

For the following exercises, find the critical points of the function by using algebraic techniques (completing the square) or by examining the form of the equation. Verify your results using the partial derivatives test. $$f(x, y)=-x^{2}-5 y^{2}+8 x-10 y-13$$

Step-by-Step Solution

Answer

The critical point is at \((4, -1)\), confirmed as a local maximum by the second derivative test.

Step 1: Rewrite the Function

Given the function, we need to rewrite it in a form that allows us to find its critical points more easily. Our function is:\[ f(x, y) = -x^2 - 5y^2 + 8x - 10y - 13. \]We aim to express it in a 'completed square' form for both variables separately.

Step 2: Complete the Square for x

Concentrate on the terms involving \(x\): \[ -x^2 + 8x. \]Complete the square: \[ -x^2 + 8x = -(x^2 - 8x) = -(x^2 - 8x + 16 - 16) = -(x-4)^2 + 16. \]

Step 3: Complete the Square for y

Now focus on the terms involving \(y\):\[ -5y^2 - 10y. \]Complete the square for this expression:\[ -5(y^2 + 2y) = -5(y^2 + 2y + 1 - 1) = -5((y+1)^2 - 1) = -5(y+1)^2 + 5. \]

Step 4: Rewrite the Full Function

Substituting back the completed squares, our function becomes:\[ f(x, y) = -(x-4)^2 + 16 - 5(y+1)^2 + 5 - 13. \]Simplify the constants:\[ f(x, y) = -(x-4)^2 - 5(y+1)^2 + 8. \]

Step 5: Identify the Critical Point

From the completed square form, the expression achieves its maximum when both squares are zero, which happens at \(x = 4\) and \(y = -1\). Thus, the critical point is \((x, y) = (4, -1).\)

Step 6: Verify Critical Point via Partial Derivatives

Calculate the partial derivatives to confirm the critical point. The partial derivative with respect to \(x\):\[ \frac{\partial f}{\partial x} = -2x + 8. \]Setting \( \frac{\partial f}{\partial x} = 0 \), we get \(-2x + 8 = 0\) leading to \(x = 4\).The partial derivative with respect to \(y\):\[ \frac{\partial f}{\partial y} = -10y - 10. \]Setting \( \frac{\partial f}{\partial y} = 0 \), we get \(-10y - 10 = 0\) leading to \(y = -1\).Both derivatives confirming the critical point \((4, -1)\).

Step 7: Second Derivative Test

To further verify, apply the second derivative test. Calculate the second partial derivatives:\(\frac{\partial^2 f}{\partial x^2} = -2\),\(\frac{\partial^2 f}{\partial y^2} = -10\),\(\frac{\partial^2 f}{\partial x \partial y} = 0\).The Hessian determinant \(D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2\):\[ D = (-2)(-10) - 0^2 = 20 > 0. \]Since \(\frac{\partial^2 f}{\partial x^2} < 0\) and \(D > 0\), this confirms a local maximum at \((x, y) = (4, -1)\).

Key Concepts

Partial Derivatives

Partial derivatives are an essential tool in calculus, especially when dealing with functions of multiple variables. Imagine partial derivatives as the way to measure how a function changes as each variable changes, while keeping the other variables constant.
The partial derivative of a function with respect to one of its variables is like a regular derivative, but we treat the other variables as constants. This gives us insight into the function's behavior in different directions.When finding critical points for a function of two variables, like our function \( f(x, y) = -x^2 - 5y^2 + 8x - 10y - 13 \), we calculate the partial derivatives with respect to both variables \( x \) and \( y \).

• Partial with respect to \( x \): Differentiate the function with respect to \( x \), treating \( y \) as a constant. This measures change along the \( x \)-axis.
• Partial with respect to \( y \): Differentiate with respect to \( y \), treating \( x \) as a constant. This measures change along the \( y \)-axis.

These partials were set to zero in our solution to find the critical point \((4, -1)\) by solving \( \frac{\partial f}{\partial x} = -2x + 8 = 0 \) and \( \frac{\partial f}{\partial y} = -10y - 10 = 0 \). Finding where these derivatives are zero helps identify potential peaks, valleys, or saddle points in the landscape drawn by the function.

Completing the Square

Completing the square is a technique used to rewrite quadratic expressions in a form that makes it easy to identify maximum or minimum values. This is especially helpful in locating critical points of functions.
In a quadratic expression like \( ax^2 + bx + c \), completing the square involves restructuring it to the form \( a(x-h)^2 + k \), where \( (h, k) \) reveals where the function has its extremum (either maximum or minimum).For our function \( f(x, y) = -x^2 - 5y^2 + 8x - 10y - 13 \), completing the square was done separately for \( x \) and \( y \).

• Step for \( x \): Start with \( -x^2 + 8x \). By factoring and completing the square, it becomes \( -(x-4)^2 + 16 \).
• Step for \( y \): Starts with \( -5y^2 - 10y \). This simplifies to \( -5(y+1)^2 + 5 \) after completing the square.

By substituting these completed squares back into the function, we arrived at \( f(x, y) = -(x-4)^2 - 5(y+1)^2 + 8 \).
This form makes it clear that the function reaches its maximum when both squares are zero, leading to the critical point \((4, -1)\). The completed square method is powerful for transforming a messy quadratic-looking function into something manageable.

Second Derivative Test

The Second Derivative Test in multivariable calculus helps verify the nature of a critical point—whether it's a local maximum, minimum, or a saddle point. This hinges on computing second-order derivatives and organizing them into a Hessian matrix.
For a function \( f(x, y) \), the Hessian consists of the second partial derivatives:

• \( \frac{\partial^2 f}{\partial x^2} \): The second partial with respect to \( x \).
• \( \frac{\partial^2 f}{\partial y^2} \): The second partial with respect to \( y \).
• \( \frac{\partial^2 f}{\partial x \partial y} \): The mixed second partial derivative.

Next, you calculate the determinant of the Hessian:\[ D = \left( \frac{\partial^2 f}{\partial x^2} \right) \left( \frac{\partial^2 f}{\partial y^2} \right) - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2 \]In our example, we have:

• \( \frac{\partial^2 f}{\partial x^2} = -2 \)
• \( \frac{\partial^2 f}{\partial y^2} = -10 \)
• \( \frac{\partial^2 f}{\partial x \partial y} = 0 \)

The determinant \( D \) is calculated as \( 20 > 0 \), confirming a local extremum. It being positive, and with \( \frac{\partial^2 f}{\partial x^2} < 0 \), identifies the point \((4, -1)\) as a local maximum. Thus, the Second Derivative Test is an efficient way to categorize a critical point's nature while considering the curvature in the neighborhood of the point.