Problem 30 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

Max: 11 at (4, 3); Min: -2 at (1, 1).

Step 1: Identify the constraint region

The constraints are given as inequalities: \(x \geq 0\), \(y \geq 0\), \(x \leq 4\), and \(y \leq 3\). These define a rectangle in the first quadrant of the xy-plane with vertices at \((0, 0), (4, 0), (4, 3), \text{ and } (0, 3)\).

Step 2: Evaluate the function at each vertex

Compute the value of the function \(f(x, y) = x^2 + y^2 - 2x - 2y\) at each of the vertices: - Vertex \((0, 0)\): \(f(0, 0) = 0^2 + 0^2 - 2 \cdot 0 - 2 \cdot 0 = 0\) - Vertex \((4, 0)\): \(f(4, 0) = 4^2 + 0^2 - 2 \cdot 4 - 2 \cdot 0 = 16 - 8 = 8\) - Vertex \((4, 3)\): \(f(4, 3) = 4^2 + 3^2 - 2 \cdot 4 - 2 \cdot 3 = 25 - 14 = 11\)- Vertex \((0, 3)\): \(f(0, 3) = 0^2 + 3^2 - 2 \cdot 0 - 2 \cdot 3 = 9 - 6 = 3\).

Step 3: Find interior critical points, if any

To find any critical points inside the region, we set the partial derivatives of \(f(x, y)\) to zero:- \(\frac{\partial f}{\partial x} = 2x - 2 = 0 \Rightarrow x = 1\)- \(\frac{\partial f}{\partial y} = 2y - 2 = 0 \Rightarrow y = 1\).The critical point \((1, 1)\) is inside the region. Calculate \(f(1, 1) = 1^2 + 1^2 - 2 \times 1 - 2 \times 1 = 2 - 4 = -2\).

Step 4: Compare and conclude

Comparing all the calculated values: - \(f(0, 0) = 0\)- \(f(4, 0) = 8\)- \(f(4, 3) = 11\)- \(f(0, 3) = 3\)- \(f(1, 1) = -2\)The absolute maximum value is \(11\) at \((4, 3)\). The absolute minimum value is \(-2\) at \((1, 1)\).

Key Concepts

Absolute Maximum and Minimum

When dealing with functions and constraints, finding the absolute maximum and minimum values is essential. These values represent the highest and lowest points the function can achieve within the defined region. To determine these points, we evaluate the function at all possible locations that might hold these extreme values.

Here are some steps to consider:

• Evaluate the function at each corner (or vertex) of the constraint region, as these are potential candidates for maximum or minimum values.
• Analyze any critical points found inside the constraint region. These are points where the function's slope is zero, indicating a potential maximum or minimum point.
• Finally, compare the values at these points and identify which is the largest (absolute maximum) and smallest (absolute minimum).

In our exercise, the absolute maximum value is 11 at the vertex (4, 3), and the absolute minimum is -2 at the critical point (1, 1). Understanding these concepts helps in optimizing functions efficiently.

Constraint Region

The constraint region is a crucial part of optimization problems. It defines the boundaries or limitations within which the function needs to be optimized. These constraints are usually given in terms of inequalities that describe a physical or logical limitation.

In our exercise, the constraint region is bounded by

• Inequalities:
• \(x \geq 0\), \(y \geq 0\)
• \(x \leq 4\), and \(y \leq 3\).

These define a rectangle in the first quadrant of the xy-plane. The vertices of this rectangle are (0, 0), (4, 0), (4, 3), and (0, 3).

Understanding the constraint region helps determine where to evaluate the function for maxima or minima. It's within this region that we calculate values at the vertices and additional critical points.

Critical Points

Critical points are where the function's derivative is zero or undefined, which often indicates a change of direction, or slope, of the function. These points are potential candidates for local maximums or minimums. However, when under constraint regions, critical points inside these regions become crucial for finding absolute extrema.

In this context:

• Compute the partial derivatives of the function with respect to each variable separately.
• Set these derivatives equal to zero to solve for the critical points.
• Check if these critical points lie within the constraint region. If they do, evaluate the function at these points as well.

In the given problem, after taking the derivatives and setting them to zero, we discovered a critical point at (1, 1). It lies within our constraint region, and evaluating the function at this point provides crucial data for determining the absolute minimum.