Chapter 12.8, Q7

Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point.

\(f(x,y) = x^2 + 3y^2 - 6y + 4xy\)

Below is AI-generated code that produces a plot of the function.

# Define the function f(x, y)
f <- function(x, y) {
  return(x^2 + 3*y^2 - 6*y + 4*x*y)
}

# Create x and y values for plotting
x <- seq(-10, 10, length.out = 100)
y <- seq(-10, 10, length.out = 100)
z <- outer(x, y, f)

# Plot the function
persp(x, y, z, theta = 30, phi = 30, expand = 0.5, col = "lightblue", border = NA,
      xlab = "x", ylab = "y", zlab = "f(x, y)",
      main = "3D plot of f(x, y) = x^2 + 3y^2 - 6y + 4xy")

Step 1: Find the first order partial derivatives for x and y, the second order derivatives.

\(\frac{\partial f}{\partial x} = 2x + 4y\)

\(\frac{\partial f}{\partial y} = 6y - 6 + 4x\)

\(\frac{\partial^2 f}{\partial x^2} = 2\)

\(\frac{\partial^2 f}{\partial y^2} = 6\)

\(\frac{\partial^2 f}{\partial y \partial x} = 4\)

Step 2: Find the critical points.

Set \(\frac{\partial f}{\partial x} = 0\)

\(2x + 4y = 0\)

\(x + 2y = 0\)

\(x = -2y\)

Set \(\frac{\partial f}{\partial y} = 0\)

\(6y - 6 + 4x = 0\)

Substitute \(x = -2y\)

\(6y - 6 + 4(-2y) = 0\)

\(6y - 6 - 8y = 0\)

\(3y -3 - 4y = 0\)

\(-y-3 = 0\)

\(y=3\)

Recall that \(x = -2y\). Substitute \(y=3\).

\(x=-2(3) = -6\)

Therefore, the only critical point is \((-6,3)\).

Step 3: Classify the critical points.

\(D = D(a,b) = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - [\frac{\partial^2 f}{\partial y \partial x}]^2\)

\(D(-6, 3) = 2 \cdot 6 - (4)^2 = 12 - 16 = -4\)

Since \(D < 0\), \((-6,3)\) is a saddle point. There are no relative minima or maxima.