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Chapter 6: Problem 26

Find the absolute maximum and minimum values of cach function, subject to the given constraints. \(f(x, y)=x^{2}+y^{2}-2 x-2 y ; \quad x \geq 0, y \geq 0, x \leq 4\) and \(y \leq 3\)

Short Answer

Expert verified
The absolute maximum value is 11 at (4,3) and the absolute minimum value is 0 at (0,0).

Step by step solution

01

- Identify the Constraint Boundaries

Identify the constraints of the given problem: 1. x 鈮� 0 2. y 鈮� 0 3. x 鈮� 4 4. y 鈮� 3. The feasible region is a rectangle with vertices at (0,0), (4,0), (4,3), and (0,3) in the xy-plane.
02

- Evaluate the Function at the Vertices

Evaluate the function at each vertex of the feasible region to find potential maxima and minima. At (0,0): \[f(0,0) = 0^2 + 0^2 - 2(0) - 2(0) = 0\] At (4,0): \[f(4,0) = 4^2 + 0^2 - 2(4) - 2(0) = 16 - 8 = 8\] At (4,3): \[f(4,3) = 4^2 + 3^2 - 2(4) - 2(3) = 16 + 9 - 8 - 6 = 11\] At (0,3): \[f(0,3) = 0^2 + 3^2 - 2(0) - 2(3) = 9 - 6 = 3\]
03

- Compare the Function Values

Compare the function values calculated at each vertex. The values are 0, 8, 11, and 3. Absolute Maximum: The highest value is 11 at the point (4,3). Absolute Minimum: The lowest value is 0 at the point (0,0).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

calculus
Calculus is a branch of mathematics that studies how things change. It helps us find the rate at which quantities change and find extreme values (like maxima and minima). In this exercise, we're using calculus to identify the absolute maximum and minimum values of a function subject to certain constraints.
The function given is \( f(x, y) = x^2 + y^2 - 2x - 2y \). Our goal is to find the highest and lowest values of this function within a specific region.
Calculus can often involve taking derivatives to find these extreme values. However, in this case, we will use a simpler method by evaluating the function at specific points.
constraint boundaries
Constraint boundaries are the limits within which we need to find the solution. They define the region in which we will search for the maximum and minimum values.
For our problem, the constraints are:
  • \( x \geq 0 \)
  • \( y \geq 0 \)
  • \( x \leq 4 \)
  • \( y \leq 3 \)
These constraints form a rectangle in the xy-plane. The rectangle's vertices are at (0,0), (4,0), (4,3), and (0,3). These corners, or vertices, are the most important parts of the feasible region to evaluate for maximum and minimum values.
feasible region
The feasible region is the space within which we find our solution. In this exercise, the feasible region is a rectangle defined by the constraint boundaries.
This rectangle is bounded by:
  • \(x \geq 0 \)
  • \(y \geq 0 \)
  • \(x \leq 4 \)
  • \(y \leq 3 \)
The feasible region is essential because our solution must lie within this area. It narrows down the possible points where we can find the maximum and minimum values.
vertex evaluation
Vertex evaluation is a method used in optimization problems to find the extreme values of a function. Since our feasible region is a rectangle, the maximum and minimum values will occur at the vertices of this rectangle.
Let's evaluate the function \(f(x, y) = x^2 + y^2 - 2x - 2y \) at each vertex:
  • At (0,0): \( f(0,0) = 0^2 + 0^2 - 2(0) - 2(0) = 0 \)
  • At (4,0): \( f(4,0) = 4^2 + 0^2 - 2(4) - 2(0) = 16 - 8 = 8 \)
  • At (4,3): \( f(4,3) = 4^2 + 3^2 - 2(4) - 2(3) = 16 + 9 - 8 - 6 = 11 \)
  • At (0,3): \( f(0,3) = 0^2 + 3^2 - 2(0) - 2(3) = 9 - 6 = 3 \)
By comparing these values, we find that the absolute maximum value is 11 at the vertex (4,3) and the absolute minimum value is 0 at the vertex (0,0). This method is straightforward and very effective in this type of problem.

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Most popular questions from this chapter

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