Problem 14 - Calculus with Concepts in Calculus

Question

Find the extreme values of \(f\) in the region described by the given inequalities. In each case assume that the extreme values exist. $$ f(x, y)=16-x^{2}-4 y^{2} ; x^{4}+2 y^{4} \leq 1 $$

Step-by-Step Solution

Answer

Maximum at \((0,0): f = 16\) and minimum on boundary points.

Step 1: Identification of the Region

The region is defined by the inequality \(x^4 + 2y^4 \leq 1\). This describes a bounded region in the \(xy\)-plane, within which we will find the extreme values of the given function \(f(x, y) = 16 - x^2 - 4y^2\).

Step 2: Determine Critical Points in the Interior

To find critical points of \(f(x, y)\), compute the partial derivatives: \(f_x = -2x\) and \(f_y = -8y\). Set both to zero: \(-2x = 0\) and \(-8y = 0\), leading to the solution \(x = 0\) and \(y = 0\). Check if this point is within the region: \(0^4 + 2(0)^4 = 0 \leq 1\).

Step 3: Analyze the Boundary

The boundary is defined by \(x^4 + 2y^4 = 1\). Use the method of Lagrange multipliers to find extreme values on the boundary. Set \(abla f = \lambda abla g\), where \(g(x, y) = x^4 + 2y^4 - 1 = 0\). Compute \(abla f = (-2x, -8y)\) and \(abla g = (4x^3, 8y^3)\). Set \(-2x = 4x^3 \lambda\) and \(-8y = 8y^3 \lambda\). For \(x eq 0\) and \(y eq 0\), simplify to find possible values of \(\lambda\) and solve for \(x\) and \(y\).

Step 4: Solve Lagrange Equations

From the equations \(-2x = 4x^3 \lambda\) and \(-8y = 8y^3 \lambda\), obtain \(\lambda = -\frac{1}{2x^2}\) and \(\lambda = -\frac{1}{y^2}\). Equate to get \(\frac{1}{x^2} = \frac{1}{y^2}\), leading to \(x^2 = y^2\). Substitute \(y = \pm x\) into \(g(x, y): x^4 + 2(\pm x)^4 = 1\). This simplifies to \(9x^4 = 1\), giving \(x = \pm \sqrt[4]{\frac{1}{9}}\). Accordingly, \(y = \pm \sqrt[4]{\frac{1}{9}}\).

Step 5: Evaluate the Function at Critical and Boundary Points

Calculate \(f\) at the interior critical point \((0,0)\): \(f(0,0) = 16\). For boundary points \((\sqrt[4]{\frac{1}{9}}, \sqrt[4]{\frac{1}{9}})\), \((-\sqrt[4]{\frac{1}{9}}, -\sqrt[4]{\frac{1}{9}})\), evaluate: \(f(\sqrt[4]{\frac{1}{9}}, \sqrt[4]{\frac{1}{9}}) = 16 - \frac{2}{3} - \frac{8}{3} = \frac{22}{3}\).

Key Concepts

Calculus

Calculus plays a fundamental role in analyzing functions to determine extreme values, which are the highest or lowest points within a specific region. In this problem, the function under consideration is \(f(x, y) = 16 - x^2 - 4y^2\). This function is defined in a two-variable real number system, making it a surface in three-dimensional space.
When working with such functions, we often seek to find both maximum and minimum points, collectively known as extreme values. Calculus offers efficient methods to find these, especially within bounded regions defined by inequalities such as \(x^4 + 2y^4 \leq 1\).

• The maximum point would be the peak of the surface where the function reaches its highest value.
• The minimum point, on the other hand, would be the lowest value within the specified region.

The task requires both identifying any critical points within the interior and examining possible extremes on the defined boundary. The application of calculus in this context begins with finding the function's derivatives to locate critical points where the slope is zero.

Lagrange multipliers

Lagrange multipliers are a powerful mathematical tool used to find the maximum and minimum values of a function subject to a constraint. In this problem, we want to find the extreme values of the function \(f(x, y)\) along the boundary defined by the constraint \(x^4 + 2y^4 = 1\).
To apply this technique, we need to set the gradient of the function \(abla f\) equal to a scalar (the multiplier \(\lambda\)) times the gradient of the constraint \(abla g\).

• The gradient \(abla f = (-2x, -8y)\) represents the direction of the steepest ascent or descent for the function \(f(x, y)\).
• The constraint function \(g(x, y)\) has a gradient \(abla g = (4x^3, 8y^3)\). Setting \(abla f = \lambda abla g\) creates a system of equations that help solve for \(x, y,\) and \(\lambda\).

This method allows for the transformation of the problem into one where both the function and its constraint can be simultaneously solved, pinpointing the border extreme values necessary in this context.

Partial derivatives

Partial derivatives are essential in multivariable calculus for analyzing functions in multiple dimensions. When dealing with a function \(f(x, y)\), a partial derivative with respect to \(x\) (denoted \(f_x\)) tells us how the function changes as \(x\) varies while keeping \(y\) constant. Similarly, \(f_y\) describes how \(f\) changes with variations in \(y\), keeping \(x\) constant.
For the function \(f(x, y) = 16 - x^2 - 4y^2\), the partial derivatives are:

• \(f_x = -2x\)
• \(f_y = -8y\)

Setting these derivatives to zero identifies critical points. In this case, solving \(-2x = 0\) and \(-8y = 0\) leads to the critical point \((0,0)\). Evaluating \(f(x, y)\) at this point determines if it is a maximum, minimum, or saddle point within the region. This evaluation is crucial as it allows us to understand the behavior of the function at key locations, helping identify solutions to the posed problem.