Problem 30 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f\) on the set \(D .\) $$\begin{array}{l}{f(x, y)=3+x y-x-2 y, \quad D \text { is the closed triangular }} \\ {\text { region with vertices }(1,0),(5,0), \text { and }(1,4)}\end{array}$$

Step-by-Step Solution

Answer

The absolute maximum is 3 at (3, 2); the absolute minimum is -2 at (5, 0).

Step 1: Identify the Boundary of the Region

The region \( D \) is defined by the vertices \((1,0), (5,0), (1,4)\). This triangular region includes the line segments: \(x = 1\) from \(y = 0\) to \(y = 4\), \(y = 0\) from \(x = 1\) to \(x = 5\), and the line that connects \((5,0)\) to \((1,4)\).

Step 2: Find the Equation of the Line from (5,0) to (1,4)

The slope of the line between \((5,0)\) and \((1,4)\) is \( \frac{4-0}{1-5} = -1 \). The equation of the line is \( y = -x + 5 \).

Step 3: Evaluate the Function at the Vertices

Calculate \( f(x, y) \) at each vertex: - \( f(1, 0) = 3 + 1 \cdot 0 - 1 - 2 \cdot 0 = 2 \)- \( f(5, 0) = 3 + 5 \cdot 0 - 5 - 2 \cdot 0 = -2 \)- \( f(1, 4) = 3 + 1 \cdot 4 - 1 - 2 \cdot 4 = 0 \)

Step 4: Evaluate the Function on the Boundary x = 1

Evaluate \( f(1, y) = 3 + 1 \cdot y - 1 - 2y = 2 - y \). Test values from \(y = 0\) to \(y = 4\): - \( f(1, 0) = 2 \), - \( f(1, 4) = -2 \). The maximum is 2 and the minimum is -2 on this segment.

Step 5: Evaluate the Function on the Boundary y = 0

Evaluate \( f(x, 0) = 3 + x \cdot 0 - x - 2 \cdot 0 = 3 - x \). Test values from \(x = 1\) to \(x = 5\):- \( f(1, 0) = 2 \),- \( f(5, 0) = -2 \).The maximum is 2 and the minimum is -2 on this segment.

Step 6: Evaluate the Function on the Line y = -x + 5

Substitute \(y = -x + 5\) into \(f(x, y)\), giving \(f(x, -x + 5) = 3 + x(-x + 5) - x - 2(-x + 5)\).- Simplify to: \(f(x, -x + 5) = 3 - x^2 + 5x - x + 2x - 10\).- Rearrange to: \(-x^2 + 6x - 7\).Find the vertex of this parabola, which is at \(x = -\frac{b}{2a} = -\frac{6}{2(-1)} = 3\).- \(y = -3 + 5 = 2\).Evaluate \(f(3, 2) = 3\).Check endpoints: \(f(5,0) = -2\) and \(f(1,4) = 0\).

Step 7: Determine the Absolute Maximum and Minimum

Compare the maximum and minimum values obtained:- Absolute maximum value is 3 at point (3,2) on the line \(y=-x+5\).- Absolute minimum value is -2 at point (5,0).

Key Concepts

Absolute Maximum and Minimum

When working with calculus optimization problems, we often need to find the absolute maximum and minimum values of a function within a given region. This means we are looking for the highest and lowest values that the function can achieve within that region. The maximum and minimum can either occur at the interior of the region or along its boundary.

To find these values, calculate the function at critical points, where the first derivative equals zero or doesn't exist, as well as at the boundary of the region. Compare all these results to determine the absolute maximum and minimum.

• Evaluate at all vertices of the region.
• Check along edges by substituting edge lines into the function.
• Find any critical points on the interior or along edges, and evaluate the function there.

These steps help ensure that all potential maximum and minimum points are considered, leading to an accurate result.

Closed Region

A closed region in math is like a locked space that includes its boundary. For a two-dimensional space, a closed region could be a shape like a triangle, rectangle, or circle. It encloses all the points inside as well as those on its boundary.

In this specific exercise, the closed region is a triangular shape defined by three points called vertices: (1,0), (5,0), and (1,4). Each line segment connecting these vertices forms the boundary of the closed region.

• Edges of the triangle: The vertical edge at x = 1, the horizontal edge at y = 0, and the diagonal edge connecting (5,0) to (1,4).
• Boundary evaluation: Integral to ensure that function values on the edges and vertices are considered.

Understanding how to deal with the boundaries helps find precise maximum and minimum values.

Multivariable Functions

Multivariable functions involve more than one variable, typically represented as f(x, y) when dealing with two variables. These functions take points from a plane and assign them a value.

The function evaluated in our exercise is a multivariable function: \[ f(x, y) = 3 + xy - x - 2y \]It considers two-dimensional input (x and y) to produce a single output value. Analyzing such functions often requires special attention when learning calculus optimization.

• Handling multiple inputs: Requires finding derivatives with respect to each variable.
• Boundary behavior: Requires testing across more than one dimension.
• Critical points: Including both interior points and points along boundaries.

Working with multivariable functions equips you with broader tools for solving real-world problems that involve complex spaces.