Problem 22 - 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

Answer: The critical point of the function is located at \((1, 2)\) and corresponds to a local minimum.

Step 1: Find the first partial derivatives of the function

We'll find the partial derivatives of the function with respect to x and y: $$ \frac{\partial f}{\partial x} = 4x^3 - 4 $$ $$ \frac{\partial f}{\partial y} = 4y^3 - 32 $$

Step 2: Find the critical points by setting the first partial derivatives to zero

Equate the first partial derivatives to zero and solve for x and y: $$ 4x^3 - 4 = 0 $$ $$ 4y^3 - 32 = 0 $$ Solving for x and y, we get: $$ x^3 = 1 \Rightarrow x = 1 $$ $$ y^3 = 8 \Rightarrow y = 2 $$ So, the critical point is at \((1, 2)\).

Step 3: Compute the second partial derivatives of the function

Now we will compute the second partial derivatives: $$ \frac{\partial^2 f}{\partial x^2} = 12x^2 $$ $$ \frac{\partial^2 f}{\partial y^2} = 12y^2 $$ $$ \frac{\partial^2 f}{\partial x\partial y} = \frac{\partial^2 f}{\partial y\partial x} = 0 $$

Step 4: Apply the Second Derivative Test

Now we need to find the determinant of the Hessian matrix using the values of the second partial derivatives evaluated at the critical point \((1, 2)\): $$ D = \begin{vmatrix} 12 \cdot 1^2 & 0 \\ 0 & 12 \cdot 2^2 \end{vmatrix} = 12^2 $$ Since both the determinant \(D\) and the second partial derivative with respect to x, \(\frac{\partial^2 f}{\partial x^2}\), are positive, we can conclude that the critical point at \((1, 2)\) corresponds to a local minimum.

Step 5: Confirm the results with a graphing utility

A graph of the function \(f(x, y) = x^4 + y^4 - 4x - 32y + 10\) confirms the presence of a local minimum at the critical point \((1, 2)\). The graph's shape around this point supports the result found using the Second Derivative Test.