Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the maximum and minimum values of \(f(x, y)=10+x+y\) on the disk \(x^{2}+y^{2} \leq 9\). Hint Parametrize the boundary by \(x=3 \cos t, y=3 \sin t, 0 \leq t \leq 2 \pi\).

Step-by-Step Solution

Answer

Max value: \(10 + 3\sqrt{2}\); Min value: \(10 - 3\sqrt{2}\).

Step 1: Identify the Function and the Constraint

The function to optimize is given as \(f(x, y) = 10 + x + y\), and the constraint is represented by the disk \(x^2 + y^2 \leq 9\). This is a circle with a radius of 3 centered at the origin.

Step 2: Evaluate at the Boundaries

To check on the boundary of the disk, we use the parametric representation \(x = 3\cos(t)\) and \(y = 3\sin(t)\), where \(0 \leq t \leq 2\pi\). Substitute these into \(f(x,y)\) to get \(f(3\cos(t), 3\sin(t)) = 10 + 3\cos(t) + 3\sin(t)\).

Step 3: Differentiate the Parametric Form

Differentiate \(g(t) = 10 + 3\cos(t) + 3\sin(t)\) with respect to \(t\). The derivative is \(g'(t) = -3\sin(t) + 3\cos(t)\).

Step 4: Find Critical Points by Setting Derivative to Zero

Solve \(-3\sin(t) + 3\cos(t) = 0\) to find critical points. Simplifying, you get \(\tan(t) = 1\), which gives \(t = \frac{\pi}{4}\) and \(t = \frac{5\pi}{4}\), as these values lie within \(0 \leq t \leq 2\pi\).

Step 5: Evaluate \(f(x, y)\) at Critical Points on Boundary

Calculate \(f(x, y)\) at \(t = \frac{\pi}{4}\) and \(t = \frac{5\pi}{4}\):- For \(t = \frac{\pi}{4}\), \(x = \frac{3\sqrt{2}}{2}\) and \(y = \frac{3\sqrt{2}}{2}\). Thus, \(f\left(\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2}\right) = 10 + 3\sqrt{2}\).- For \(t = \frac{5\pi}{4}\), \(x = -\frac{3\sqrt{2}}{2}\) and \(y = -\frac{3\sqrt{2}}{2}\). Thus, \(f\left(-\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2}\right) = 10 - 3\sqrt{2}\).

Step 6: Evaluate \(f(x, y)\) at the Center of the Disk

Since the center \((0, 0)\) is inside the disk, evaluate \(f(0, 0) = 10\).

Step 7: Compare Values to Determine the Maximum and Minimum

Compare the obtained values:- At the center: \(f(0, 0) = 10\)- At \(t = \frac{\pi}{4}\): \(f\left(\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2}\right) = 10 + 3\sqrt{2}\)- At \(t = \frac{5\pi}{4}\): \(f\left(-\frac{3\sqrt{2}}{2}, -\frac{3\sqrt{2}}{2}\right) = 10 - 3\sqrt{2}\)Thus, the maximum value is \(10 + 3\sqrt{2}\) and the minimum value is \(10 - 3\sqrt{2}\).

Key Concepts

Parametrization

When dealing with optimization problems in multivariable calculus, especially those involving constrained regions like circles, parametrization becomes an incredibly useful tool. In this exercise, we aim to find maxima and minima of a function defined over a disk, specifically where the constraint is a circle with radius 3 centered at the origin. The simplest way to describe this circle is by using a parametric equation:

• Set x as a function of cosine: \( x = 3\cos(t) \)
• Set y as a function of sine: \( y = 3\sin(t) \)

These equations trace every point on the circle as t varies from 0 to \( 2\pi \). Parametrization effectively reduces the number of variables, simplifying the process of evaluating the function along the boundary. This technique is often the first step in tackling boundary optimization problems, allowing a complex 2D domain to be addressed through a single variable.

Optimization

Optimization in multivariable calculus usually involves finding where a function reaches its maximum or minimum values. In this case, we're looking at a function, \( f(x, y) = 10 + x + y \), and trying to determine its extreme values over a disk region. To effectively solve these problems:

• First, you should understand the function's domain, which is defined by the constraint \( x^2 + y^2 \leq 9 \). This outlines a circle of radius 3.
• Then, evaluate the function not just inside this circle, but especially along its boundary, as these are often locations where extrema occur.

The concept of optimization here involves evaluating the function at key points, including the parametric boundary and any potential critical points, to ascertain the function’s extremes.

Critical Points

Critical points are essential in identifying the potential maxima or minima within a region. They are points within the domain where the derivative is zero or undefined. In the context of our function and its domain:

• Convert the original multivariable function into a single variable by substituting the parametrized equations.
• This gives a new function, \( g(t) = 10 + 3\cos(t) + 3\sin(t) \), derived by plugging the parametric forms into the function, \( f(x, y) \).
• Find the derivative, \( g'(t) = -3\sin(t) + 3\cos(t) \), and set it to zero to solve for critical points.

Upon solving \( -3\sin(t) + 3\cos(t) = 0 \), we find critical points at specific values of t, namely \( t = \frac{\pi}{4} \) and \( t = \frac{5\pi}{4} \). These are potential candidates for maximum or minimum function values.

Boundary Evaluation

Boundary evaluation is crucial when identifying the extremal values of a function over a constrained region. For the exercise at hand:

• We leverage the parametrization to evaluate the function specifically on the boundary, where \( x = 3\cos(t) \) and \( y = 3\sin(t) \).
• By substituting these into our function \( f(x, y) \), we get a boundary-specific evaluation: \( f(3\cos(t), 3\sin(t)) = 10 + 3\cos(t) + 3\sin(t) \).
• Then, plug in the critical values of t obtained earlier, and compute \( f \) at these points to see if they represent the maximum or minimum values.

The process involves comparing the function’s value at these critical points to that at other key points, like the center of the disk, to identify the true extrema. Conclusively, the highest and lowest values observed indicate the function’s maximum and minimum over the given domain.