Problem 23 - Calculus Early Transcendentals

Question

Find the critical points of the following functions. Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, local minimum, or saddle point. Confirm your results using a graphing utility. $$f(x, y)=x^{4}+y^{4}-4 x-32 y+10$$

Step-by-Step Solution

Answer

Question: Based on the given function \(f(x,y) = x^4 + y^4 - 4x - 32y + 10\), determine the nature of the critical point found at \((1,2)\). Answer: The critical point \((1,2)\) is a local minimum.

Step 1: Find the first-order partial derivatives

To find the critical points of the function, we first need to find its first-order partial derivatives: $$\frac{\partial f}{\partial x} = 4x^3 - 4$$ $$\frac{\partial f}{\partial y} = 4y^3 - 32$$

Step 2: Find the critical points

We set both partial derivatives to zero and solve the system of equations to find the critical points: $$4x^3 - 4 = 0$$ $$4y^3 - 32 = 0$$ From the first equation, we have: $$x^3 = 1 \Rightarrow x = 1$$ From the second equation, we have: $$y^3 = 8 \Rightarrow y = 2$$ So, there is only one critical point, which is \((1,2)\).

Step 3: Calculate the second-order mixed partial derivatives

Now we need to find the second-order mixed partial derivatives of the function: $$\frac{\partial^2 f}{\partial x^2} = 12x^2$$ $$\frac{\partial^2 f}{\partial y^2} = 12y^2$$ $$\frac{\partial^2 f}{\partial x \partial y} = 0$$

Step 4: Apply the Second Derivative Test

We need to compute the Discriminant, D, to determine whether the critical point is a local maximum, local minimum, or saddle point: $$D(x,y) = \left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2$$ Substituting \((1,2)\) into the expressions for the second-order partial derivatives, we get: $$\frac{\partial^2 f}{\partial x^2}(1,2) = 12(1)^2 = 12$$ $$\frac{\partial^2 f}{\partial y^2}(1,2) = 12(2)^2 = 48$$ $$\frac{\partial^2 f}{\partial x \partial y}(1,2) = 0$$ Now, compute the Discriminant: $$D(1,2) = (12)(48) - (0)^2 = 576$$ Since \(D(1,2) > 0\) and \(\frac{\partial^2 f}{\partial x^2}(1,2) > 0\), we conclude that there is a local minimum at the critical point \((1,2)\).

Step 5: Confirm the results using a graphing utility

Using a graphing utility or software, plot the function \(f(x,y) = x^4 + y^4 - 4x - 32y + 10\). You will notice that the critical point \((1,2)\) corresponds to a local minimum, confirming our earlier conclusion.

Key Concepts

Critical Points

Critical points are essential in calculus, especially when analyzing multivariable functions. A critical point for a function of two variables, like our function \(f(x,y)\), is a point \((x_0,y_0)\) where the first-order partial derivatives of the function are zero.

This means that both the rate of change of the function with respect to \(x\) and \(y\) are zero at the critical point. Finding these points involves setting the partial derivatives equal to zero and solving the resulting system of equations.

• For the given function \(f(x,y) = x^4 + y^4 - 4x - 32y + 10\), we calculated:
  • \(\frac{\partial f}{\partial x} = 4x^3 - 4\)
  • \(\frac{\partial f}{\partial y} = 4y^3 - 32\)

Solving \(4x^3 - 4 = 0\) and \(4y^3 - 32 = 0\), we determined the critical point is \((1,2)\).
A critical point does not always mean there is a local minimum or maximum, which is why further investigation with the Second Derivative Test is necessary.

Partial Derivatives

Partial derivatives are like regular derivatives but applied to multivariable functions. They represent the rate of change of the function with respect to one variable while keeping other variables constant.

For a function \(f(x,y)\), the partial derivative \(\frac{\partial f}{\partial x}\) measures how \(f\) changes as \(x\) changes, while \(y\) is fixed. Similarly, \(\frac{\partial f}{\partial y}\) measures the change in \(f\) as \(y\) changes, with \(x\) fixed.

• In our example, the partial derivatives were:
  • \(\frac{\partial f}{\partial x} = 4x^3 - 4\)
  • \(\frac{\partial f}{\partial y} = 4y^3 - 32\)

These derivatives help us find critical points by determining where these rates of change are zero. The understanding of partial derivatives is crucial in the broader field of multivariable calculus, allowing us to explore more complex surfaces and their behaviors.

Local Minimum

A local minimum is a point where the function value is smaller than or equal to nearby points. In the context of multivariable functions, identifying a local minimum involves both critical points and the Second Derivative Test.

After finding a critical point, we use the second derivatives to assess curvature around that point.

• The Second Derivative Test calculates the Discriminant \(D\):
  \[ D(x,y) = \left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 \]
• In our example:
  • \(\frac{\partial^2 f}{\partial x^2}(1,2) = 12x^2 = 12\)
  • \(\frac{\partial^2 f}{\partial y^2}(1,2) = 12y^2 = 48\)
  • \(\frac{\partial^2 f}{\partial x \partial y}(1,2) = 0\)
  • The Discriminant \(D(1,2) = 576\).
• Since \(D(1,2) > 0\) and \(\frac{\partial^2 f}{\partial x^2} > 0\), there is a local minimum at \((1,2)\).

This calculation confirms that \(f(x,y)\) has a local minimum at the critical point \((1,2)\). This analysis helps us understand function behavior and graphs, confirming results visually when plotted.