Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the maximum and minimum of the function f over the closed and bounded set \(S .\) Use the methods of Section \(12.8\) to find the maximum and minimum on the the interior of \(S ;\) then use Lagrange multipliers to find the maximum and minimum over the boundary of \(S .\) \(f(x, y)=(1+x+y)^{2} ; S=\left\\{(x, y): \frac{x^{2}}{4}+\frac{y^{2}}{16} \leq 1\right\\}\)

Step-by-Step Solution

Answer

Maximum is 16; minimum is 0.

Step 1: Identify the Domain

The set \( S \) is a closed and bounded set, described by the equation \( \frac{x^2}{4} + \frac{y^2}{16} \leq 1 \), an ellipse centered at the origin with semi-major axis 4 along the y-axis and semi-minor axis 2 along the x-axis.

Step 2: Find Critical Points in the Interior

To find critical points of \( f(x, y) = (1+x+y)^2 \) on the interior of \( S \), we compute the partial derivatives and set them to zero: \( f_x = 2(1+x+y) \) and \( f_y = 2(1+x+y) \). Setting these equal to zero, we have \( 1+x+y = 0 \), yielding a line \( x + y = -1 \) in the interior of \( S \).

Step 3: Evaluate Function at Interior Critical Points

Since the equation \( 1+x+y = 0 \) represents a line, finding a specific critical point means considering points that satisfy both the ellipse definition and this line. Any intersection points need to be evaluated against boundary constraints. Typically, such intersections are boundary points since the line goes through the center.

Step 4: Apply the Method of Lagrange Multipliers

To find extrema on the boundary, use the constraint \( g(x, y) = \frac{x^2}{4} + \frac{y^2}{16} - 1 = 0 \). Construct the Lagrangian: \( \mathcal{L}(x, y, \lambda) = (1+x+y)^2 + \lambda \left( \frac{x^2}{4} + \frac{y^2}{16} - 1 \right) \).

Step 5: Set Gradient of Lagrangian to Zero

Differentiate the Lagrangian with respect to \( x \), \( y \), and \( \lambda \):\( abla \mathcal{L} = (2(1+x+y) + \lambda \cdot \frac{x}{2}, 2(1+x+y) + \lambda \cdot \frac{y}{8}, \frac{x^2}{4} + \frac{y^2}{16} - 1) \).Set equations to zero to solve for critical points across the boundary.

Step 6: Solve for Boundary Critical Points

Solving the system of equations from the previous step: \( 2(1+x+y) + \lambda \cdot \frac{x}{2} = 0 \), \( 2(1+x+y) + \lambda \cdot \frac{y}{8} = 0 \), \( \frac{x^2}{4} + \frac{y^2}{16} = 1 \) gives possible values of \( x \), \( y \). Evaluate \( f(x, y) \) for these values.

Step 7: Evaluate at Possible Extremes

Check the values of the function \( f \) at the determined boundary critical points and compare these to the values found over the potential interior critical points.

Key Concepts

Elliptical Constraints

Elliptical constraints limit the domain of a function into a specific shape, in this case, an ellipse. The given set \( S = \left\{(x, y): \frac{x^2}{4} + \frac{y^2}{16} \leq 1\right\}\) describes an ellipse, which acts as the boundary for the problem. This ellipse is centered at the origin, with a semi-major axis length of 4 along the y-axis and a semi-minor axis of 2 along the x-axis.

When dealing with such constraints, we must evaluate both the interior of the ellipse and its boundary to find all possible extrema of the function. The constraint defines how far \( x \) and \( y \) can move while staying within the bounds set by the ellipse. This means the function \( f(x, y) = (1+x+y)^2 \) is only evaluated where \( x \) and \( y \) satisfy this constraint, ensuring all points considered are within the elliptical shape.

To evaluate on the boundary, a method like Lagrange multipliers must be used, which takes the constraint into account directly during the search for extrema.

Critical Points

Critical points occur where a function's rate of change is zero; these are the points where local maximum, minimum, or saddle points could possibly occur. In this exercise, critical points for the function \( f(x, y) = (1+x+y)^2 \) are found where both partial derivatives are zero.

For \( f(x, y) \), the partial derivatives are \( f_x = 2(1+x+y) \) and \( f_y = 2(1+x+y) \). Setting these to zero results in \( 1+x+y = 0 \). This yields a line of critical points, \( x + y = -1 \), that must be evaluated within the limits of our elliptical constraint.

It's important to understand that finding critical points within the ellipse involves confirming these potential points do not fall outside of the set \( S \). If they lie within the boundary defined by the ellipse, they should next be evaluated to determine the nature of the potential extrema they represent.

Partial Derivatives

Partial derivatives are essential in identifying points where a function might reach a local maximum or minimum. They represent the rate of change of a function with respect to one of its variables, holding the other constant. In our function \( f(x, y) = (1+x+y)^2 \), the partial derivatives are computed as follows:

- \( f_x = 2(1+x+y) \) for the rate of change with respect to \( x \).
- \( f_y = 2(1+x+y) \) for the rate of change with respect to \( y \).

These equations are solved simultaneously to locate potential critical points. The aim is to find where both these derivatives equal zero, signifying a flat point for the function in the surface it forms, representing a possible extremum.

Recognizing the significance of partial derivatives helps in understanding how a function behaves locally, especially when seeking out either maximum or minimum values within boundaries like an ellipse in a constrained optimization problem.

Extrema Evaluation

Once you identify potential critical points within and on the boundary of the elliptical constraint, the next step is to evaluate the extrema. You must check values at each critical point to discern which represent local maxima, minima, or saddle points.

On the interior of the ellipse, you have already narrowed down possibilities by setting \( 1+x+y = 0 \) and confirming these against the elliptical constraint.
On the boundary, applying Lagrange multipliers incorporates the constraint directly into the evaluation process. The Lagrangian \( \mathcal{L}(x, y, \lambda) = (1+x+y)^2 + \lambda \left( \frac{x^2}{4} + \frac{y^2}{16} - 1 \right) \) is used to derive conditions under which extrema can exist. By solving the resulting system of equations derived from setting the gradient of the Lagrangian to zero, you find the critical points on the boundary.

Finally, evaluate the function \( f(x, y) \) at all these critical points to ascertain the highest and lowest values it takes. These represent the maximum and minimum values of \( f \) over the set \( S \), ensuring that all possibilities, whether on the interior or boundary of the ellipse, are considered.