Problem 15 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, a saddle point, or none of these. $$f(x, y)=400-3 x^{2}-4 x+2 x y-5 y^{2}+48 y$$

Step-by-Step Solution

Answer

The critical point is (1, 5), and it is a local maximum.

Step 1: Find the First Partial Derivatives

To find the critical points, we first need to determine where the gradient of the function is zero. Calculate the partial derivatives with respect to both \(x\) and \(y\). \[\frac{\partial f}{\partial x} = -6x + 2y - 4,\]\[\frac{\partial f}{\partial y} = 2x - 10y + 48.\]

Step 2: Solve the System of Equations

Set the first partial derivatives equal to zero to find the critical points:\(-6x + 2y - 4 = 0\)\(2x - 10y + 48 = 0\)Solve this system of linear equations. From the first equation, we have: \[2y = 6x + 4 \Rightarrow y = 3x + 2\]Substitute \(y = 3x + 2\) into the second equation:\[2x - 10(3x + 2) + 48 = 0\]\[2x - 30x - 20 + 48 = 0\]\[-28x + 28 = 0\]\[x = 1\]Substitute \(x = 1\) back into \(y = 3x + 2\):\[y = 3(1) + 2 = 5\]So the critical point is \((1, 5)\).

Step 3: Calculate the Second Partial Derivatives

Compute the second partial derivatives to use in the Hessian determination for classification of critical points:\[\frac{\partial^2 f}{\partial x^2} = -6,\]\[\frac{\partial^2 f}{\partial y^2} = -10,\]\[\frac{\partial^2 f}{\partial x \partial y} = 2.\]

Step 4: Construct the Hessian Matrix

The Hessian matrix \(H\) is calculated at the critical point \((1, 5)\):\[H = \begin{bmatrix}-6 & 2 \2 & -10\end{bmatrix}\]

Step 5: Determine the Nature of the Critical Point

Find the determinant of the Hessian matrix:\[D = (-6)(-10) - (2)(2) = 60 - 4 = 56\]Since \(D > 0\) and \(\frac{\partial^2 f}{\partial x^2} = -6 < 0\), the critical point \((1, 5)\) is a local maximum.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus and are especially useful when we are working with functions of multiple variables. Imagine you have a 3D surface as a function of two variables, say \(x\) and \(y\). The partial derivative with respect to \(x\) tells us how the function is changing as \(x\) changes, while keeping \(y\) constant. Similarly, the partial derivative with respect to \(y\) gives the rate of change as \(y\) varies, with \(x\) held steady.Calculating partial derivatives involves taking the derivative with respect to one of the variables at a time:

• For \(f(x, y)=400-3x^{2}-4x+2xy-5y^{2}+48y\):
• The partial derivative with respect to \(x\) is \(-6x + 2y - 4\).
• The partial derivative with respect to \(y\) is \(2x - 10y + 48\).

These derivatives help us find critical points by setting them equal to zero, which locates where the slope is flat—that is where the function might have a local maximum, minimum, or saddle point.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives that provides critical insight into the concavity and convexity of a function. In two-variable calculus, it's instrumental in determining the nature of critical points—a step beyond simply finding them.To construct a Hessian matrix for a function \(f(x, y)\):

• You find the second partial derivatives \(\frac{\partial^2 f}{\partial x^2}\), \(\frac{\partial^2 f}{\partial y^2}\), and \(\frac{\partial^2 f}{\partial x \partial y}\).

For example, given \(f(x, y)=400-3x^{2}-4x+2xy-5y^{2}+48y\), we calculate:

• \[ \frac{\partial^2 f}{\partial x^2} = -6 \]
• \[ \frac{\partial^2 f}{\partial y^2} = -10 \]
• \[ \frac{\partial^2 f}{\partial x \partial y} = 2 \]

Thus, the Hessian matrix \(H\) is:\[\begin{bmatrix}-6 & 2\ 2 & -10\end{bmatrix}\] which helps determine if a critical point is a local extremum or a saddle point by examining its determinant.

Local Maximum

Understanding what a local maximum is helps us comprehend the significance of critical points. A local maximum is a point where the function takes a peak value compared to nearby points. It’s like a small hill in a landscape where the top is higher than any immediate surroundings.To determine if a critical point is a local maximum when functions involve several variables, the Hessian matrix is utilized:

• Calculate the determinant \(D\) of the Hessian matrix.
• If \(D > 0\) and the second partial derivative \(\frac{\partial^2 f}{\partial x^2} < 0\), it is classified as a local maximum.

In this exercise, at critical point \((1, 5)\), the determinant \(D = 56 > 0\) and \(\frac{\partial^2 f}{\partial x^2} = -6 < 0\), confirming it as a local maximum. This means at \((1, 5)\), the function \(f(x, y)\) achieves a maximum within a small neighborhood of \((x, y)\).

System of Equations

Systems of equations are essential for solving problems involving multiple unknowns, especially in calculus when searching for critical points using state-of-the-art mathematical tools like partial derivatives. They look intimidating, but they are just sets of equations solved together.For a function \(f(x, y)\), after computing the partial derivatives, we set them to zero to form a system of equations to find critical points. For example:

• \(-6x + 2y - 4 = 0\)
• \(2x - 10y + 48 = 0\)

By solving these equations, we find the values of \(x\) and \(y\) that make both partial derivatives zero, indicating flatness on the surface—potential critical points. Substitute one equation into the other: solve the first equation for \(y\), plug it into the second and solve for \(x\).This results in \((x,y)=(1,5)\) as a solution, revealing the critical point crucial in analyzing the function's behavior. By learning this, you enable yourself to explore many interesting applications in mathematics and the sciences.