Problem 44 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of the following functions on the given set \(R\). $$f(x, y)=2 x^{2}+y^{2} ; R=\left\\{(x, y): x^{2}+y^{2} \leq 16\right\\}$$

Step-by-Step Solution

Answer

Answer: The absolute maximum value of the function on the given region is 32, and the absolute minimum value is 16.

Step 1: Compute the gradient of the function and the constraint

Compute the gradient of the given function \(f(x, y) = 2x^2 + y^2\) and the constraint equation \(g(x, y) = x^2 + y^2 - 16\). For \(f(x, y)\), $$\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (4x, 2y)$$ For \(g(x, y)\), $$\nabla g(x, y) = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right) = (2x, 2y)$$

Step 2: Apply the method of Lagrange multipliers

Applying the method of Lagrange multipliers, we have: $$\nabla f(x, y) = \lambda \nabla g(x, y)$$ So, \((4x, 2y) = \lambda (2x, 2y)\). Divide both sides by 2, $$(2x, y) = \lambda (x, y)$$

Step 3: Solve for critical points

Since \((2x, y) = \lambda (x, y)\), it means that either \(\lambda = 2\) or \(y=0\). If \(\lambda = 2\), then we have \(x^2 + y^2 = 16\). If \(y=0\), then \(x^2 = 16\), so \(x = \pm 4\). We then have the following cases: 1. \(\lambda = 2\): We get the equation \(x^2 + y^2 = 16\). Here, we consider the circle with radius 4, which represents the boundary of our region \(R\). 2. \(y = 0\): We get \(x = \pm 4\). These two points are on the boundary of the region and lie on the x-axis.

Step 4: Evaluate the function at the critical points and find the maximum and minimum values

Now, we need to evaluate the function at these critical points. Case 1: \(\lambda = 2\) (Boundary of region \(R\)) Parametrize the circle using polar coordinates: \(x = 4\cos(\theta)\) and \(y = 4\sin(\theta)\) Then, the function becomes: $$f(\theta) = 2(4\cos\theta)^2 + (4\sin\theta)^2 = 32\cos^2\theta + 16\sin^2\theta$$ Its maximum and minimum will occur at the endpoints of the region and at the points where the derivative is zero: $$\frac{df}{d\theta} = -64\cos\theta\sin\theta + 32\sin\theta\cos\theta = 0$$ $$\Rightarrow \sin\theta\cos\theta = 0 \Rightarrow \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$$ Evaluate the function at these points: $$f(0) = 32, f\left(\frac{\pi}{2}\right) = 16, f(\pi) = 32, f\left(\frac{3\pi}{2}\right) = 16$$ Case 2: \(y = 0\) (Points \(x = \pm 4\) on the x-axis) Evaluate the function at these points: $$f(4, 0) = 32, f(-4, 0) = 32$$

Step 5: Determine the absolute maximum and minimum values

Compare the values from both cases: - Absolute maximum: \(f(x, y) = 32\) at points \((\pm 4, 0)\) and on the boundary of region \(R\) where \(\theta = 0\) and \(\theta = \pi\). - Absolute minimum: \(f(x, y) = 16\) at points \((0, \pm 4)\) and on the boundary of region \(R\) where \(\theta = \frac{\pi}{2}\) and \(\theta = \frac{3\pi}{2}\). Therefore, the absolute maximum value of \(f(x, y)\) on the given region \(R\) is 32, and the absolute minimum value is 16.

Key Concepts

Gradient

The gradient is a vector that points in the direction of the steepest increase of a function. For a function of two variables, like our exercise with function \(f(x, y) = 2x^2 + y^2\), its gradient is a vector consisting of the partial derivatives with respect to each variable. In this case, the gradient \(abla f(x, y)\) is \((4x, 2y)\). Each component of the gradient tells us how much the function increases per unit increase in \(x\) or \(y\).

• \(\frac{\partial f}{\partial x} = 4x\) means if \(x\) changes, the function increases four times the amount of \(x\) change.
• \(\frac{\partial f}{\partial y} = 2y\) means if \(y\) changes, the function increases twice the amount of \(y\) change.

The gradient is crucial in optimization problems, helping to identify the direction of maximum increase and used in identifying critical points where these increases are zero.

Critical Points

Critical points are where the function's output doesn't change, meaning the gradient is zero or undefined. In Lagrange multipliers, you solve for where the gradients of your function and constraint are parallel, indicating they "balance" against each other. So, for our exercise, we use the gradients: \(abla f(x, y) = (4x, 2y)\) and \(abla g(x, y) = (2x, 2y)\).With Lagrange multipliers, you set \(abla f(x, y) = \lambda abla g(x, y)\), giving \((4x, 2y) = \lambda (2x, 2y)\). Solving this gives conditions for critical points. For instance:

• When \(\lambda = 2\), the expression simplifies, indicating a critical point along the constraint boundary (\(x^2 + y^2 = 16\)).
• When \(y = 0\), the function isolates \(x\) to \(\pm 4\), giving points on the x-axis where the function might have extreme values.

Critical points help identify where potential maxima and minima occur within the domain restricted by the constraint.

Polar Coordinates

Polar coordinates offer a convenient way to describe locations on a circle, especially useful with circular constraints like \(x^2 + y^2 = 16\), forming part of our exercise.To convert to polar coordinates, use:

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\)

For a circle of radius 4 as in our problem, this becomes \(x = 4\cos(\theta)\) and \(y = 4\sin(\theta)\). This change simplifies exploring positions along the boundary, making calculus and maximization cleaner, as we express the function \(f(x,y)\) in terms of \(\theta\).Evaluating \(f(\theta) = 32\cos^2(\theta) + 16\sin^2(\theta)\) at strategic \(\theta\) values helps to locate maximum and minimum efficiently. Polar coordinates merge circular geometry with traditional algebra.

Absolute Maximum and Minimum

Finding absolute maxima and minima means determining the highest and lowest values of a function over a specified region. In optimization problems, especially those with constraints, it involves checking the value of the function at critical points and along the boundary.In our case, the function \(f(x, y) = 2x^2 + y^2\) on the region \(R: x^2 + y^2 \leq 16\) has critical values calculated both at points where the function gradient aligns with that of the constraint and along the boundary defined by \(x^2 + y^2 = 16\).After substituting values from both sets of critical points (from polar angles and direct calculation when \(y = 0\)) into \(f(x,y)\), we found:

• Absolute Maximum: 32 at points \((\pm 4, 0)\) and when \(\theta = 0, \pi\).
• Absolute Minimum: 16 at points \((0, \pm 4)\) and when \(\theta = \frac{\pi}{2}, \frac{3\pi}{2}\).

Recognizing absolute extrema allows a full understanding of a function's behavior across its feasible domain.