Problem 34 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f\) on the set \(D .\) $$f(x, y)=x y^{2}, \quad D=\\{(x, y) | x \geqslant 0, y \geqslant 0, x^{2}+y^{2} \leqslant 3\\}$$

Step-by-Step Solution

Answer

The absolute maximum value is \(\frac{9}{8}\) and the minimum value is 0.

Step 1: Identify the Domain

The function given is \(f(x, y) = x y^2\). We need to find the absolute maximum and minimum of this function within the set \(D\), which is a disk in the first quadrant (since \(x \geq 0\) and \(y \geq 0\)) and has a radius of \(\sqrt{3}\).

Step 2: Compute the Gradient

Compute the gradient of \(f\): \[abla f = \left( y^2, 2xy \right)\]Set each component of the gradient equal to zero to find critical points:- \(y^2 = 0\) implies \(y = 0\).- \(2xy = 0\) implies \(x = 0\) or \(y = 0\).Therefore, the critical point is at \((0, 0)\).

Step 3: Evaluate Critical Points

Evaluate \(f\) at critical point \((0, 0)\): \[f(0, 0) = 0 \times 0^2 = 0\]

Step 4: Evaluate Boundary

The boundary is defined by the equation \(x^2 + y^2 = 3\) with other constraints \(x \geq 0\), \(y \geq 0\). Parametrize the boundary using polar coordinates: - Let \(x = \sqrt{3} \cos \theta\) and \(y = \sqrt{3} \sin \theta\).Substitute these into \(f\):\[f(\sqrt{3} \cos \theta, \sqrt{3} \sin \theta) = (\sqrt{3} \cos \theta)(\sqrt{3} \sin \theta)^2 = 3 \cos \theta \sin^2 \theta\]Let \( g(\theta) = 3 \cos \theta \sin^2 \theta \) and analyze this function for \(0 \leq \theta \leq \frac{\pi}{2}\).

Step 5: Find Extrema on Boundary

To find the extrema, compute the derivative \(g'(\theta)\):\[g'(\theta) = 3 \cdot (-\sin \theta) \sin^2 \theta + 6 \cos \theta \sin \theta \cdot \cos \theta\]Setting \(g'(\theta) = 0\) gives:\[- \sin^3 \theta + 2 \sin \theta \cos^2 \theta = 0 \Rightarrow \sin \theta (2 \cos^2 \theta - \sin^2 \theta) = 0\]Thus \(\sin \theta = 0\) or \(2 \cos^2 \theta = \sin^2 \theta\).

Step 6: Evaluate Boundary Critical Points

- For \(\sin \theta = 0\), \(\theta = 0\), then \(g(0) = 0\).- For \(2 \cos^2 \theta = \sin^2 \theta\), solving gives \(\theta = \frac{\pi}{3}\) leading to:\[f(\sqrt{3}/2, \sqrt{3}\sqrt{3}/2) = 3 \left(\frac{1}{2}\right) \left(\frac{3}{4}\right) = \frac{9}{8}\]Also evaluate at the endpoints:- \(\theta = \frac{\pi}{2}\): \(g(\frac{\pi}{2}) = 0\).

Step 7: Conclusion

The function \(f\) within the given domain attains:- Minimum value of \(0\) at \((0,0)\), boundary points with \(\theta = 0, \frac{\pi}{2}\).- Maximum value of \(\frac{9}{8}\) at the boundary point corresponding to \(\theta = \frac{\pi}{3}\).

Key Concepts

Gradient

The concept of the gradient is quite crucial when dealing with multivariable functions. For a function of two variables, like in our problem, the gradient is a vector that points in the direction of the greatest rate of increase of the function. In mathematical terms, if you have a function \( f(x, y) \), the gradient \( abla f \) is given by \(( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} )\).

• The "\( \frac{\partial f}{\partial x} \)" component shows the rate of change of \( f \) as \( x \) changes, holding \( y \) constant.
• The "\( \frac{\partial f}{\partial y} \)" component shows the rate of change as \( y \) changes, holding \( x \) constant.

In our problem, the gradient \( abla f = (y^2, 2xy) \) is computed by taking the partial derivatives of the function \( f(x, y) = x y^2 \). This is useful for locating critical points of the function where potential extrema (maximum or minimum values) may occur.

Critical Points

Critical points are vital when determining extrema, as they are positions where the gradient is zero. This means the rate of change in every direction is zero, usually indicating a peak or trough. In functions of multiple variables, you set each component of the gradient to zero and solve for \( x \) and \( y \) to find these points.In the exercise:

• Setting \( y^2 = 0 \) gives \( y = 0 \).
• Setting \( 2xy = 0 \) gives either \( x = 0 \) or \( y = 0 \).

After solving these equations, you discover the only critical point in the domain is at \((0,0)\). This point is then evaluated to determine the function's behavior there.

Boundary Analysis

Analyzing the boundary of the domain is necessary to ensure all possible extrema have been considered. In many multivariable calculus problems, critical points alone might not account for all extrema. Here, the boundary is described by the equation \( x^2 + y^2 = 3 \) in the domain \( D \), which represents a section of a disk.Consider:

• Checking both on and inside this boundary helps find potential extrema that could occur on the edge of the domain.
• The boundary itself is analyzed, considering the constraints \( x \geq 0 \) and \( y \geq 0 \).

It is essential to understand the context of these constraints and apply them when searching for absolute maxima or minima.

Polar Coordinates

Using polar coordinates is a clever technique to simplify problems with circular or radial symmetry. This coordinate system is based on a radius and angle rather than \( x \) and \( y \), making it ideal for dealing with circular domains.For any point \((x, y)\), convert to polar using:

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

In our exercise:

• Setting \( r = \sqrt{3} \), the boundary becomes \( \theta \) varying in \([0, \pi/2]\) due to the domain's constraints.
• This simplifies the function to solve for extremal values along the curve by examining how the function changes as \( \theta \) varies.

Polar coordinates are powerful for parametrizing curves and boundaries, leading to simpler derivatives and easier boundary analysis.