Problem 32 - Calculus and its applications

Question

Find the absolute maximum and minimum values of each function, subject to the given constraints. $$ \begin{array}{l} h(x, y)=x^{2}+y^{2}-4 x-2 y+1 ; \quad x \geq 0, y \geq 0 \\ \text { and } x+2 y \leq 5 \end{array} $$

Step-by-Step Solution

Answer

The absolute maximum is 6 at (5, 0) and the absolute minimum is -4 at (2, 1).

Step 1: Identify the Objective Function and Constraints

The objective function given is \( h(x, y) = x^2 + y^2 - 4x - 2y + 1 \), which we need to optimize. The constraints are: \( x \geq 0 \), \( y \geq 0 \), and \( x + 2y \leq 5 \). We must find the absolute extreme values of the function under these conditions.

Step 2: Determine Feasible Region

The feasible region is bounded by the constraints \( x \geq 0 \), \( y \geq 0 \), and the line \( x + 2y = 5 \). First determine intercepts: when \( x = 0 \), \( y = \frac{5}{2} \); and when \( y = 0 \), \( x = 5 \). The region is bounded by these intercepts and the axes.

Step 3: Evaluate Critical Points in Interior

Take the partial derivatives: \( \frac{\partial h}{\partial x} = 2x - 4 \) and \( \frac{\partial h}{\partial y} = 2y - 2 \). Set these to zero: \( 2x - 4 = 0 \) gives \( x = 2 \); \( 2y - 2 = 0 \) gives \( y = 1 \). The point \( (2, 1) \) lies within the feasible region.

Step 4: Evaluate Boundary Points

Evaluate the function at the boundary lines. The boundaries are \( x = 0 \), \( y = 0 \), and \( x + 2y = 5 \). On the line \( x + 2y = 5 \), assume \( y = t \), then \( x = 5 - 2t \). Evaluate \( h(5-2t, t) \) to find the range on this line.

Step 5: Check Corners of the Feasible Region

The corners of the region are the intersection points of the constraints: \( (0, 0) \), \( (5, 0) \), and \( (0, \frac{5}{2}) \). Evaluate \( h(x, y) \) at these points.

Step 6: Compare Functional Values

Calculate \( h(x, y) \) for each point: \((2, 1), (0, 0), (5, 0), (0, \frac{5}{2})\). Compare to identify the maximum and minimum values of \( h(x, y) \).

Step 7: Identify Absolute Extrema

Evaluate: \( h(2, 1) = -4 \), \( h(0, 0) = 1 \), \( h(5, 0) = 6 \), \( h(0, \frac{5}{2}) = 5.25 \). Absolute minimum is \(-4\) at point \((2, 1)\); absolute maximum is \(6\) at point \((5, 0)\).

Key Concepts

Feasible Region

In any optimization problem, the feasible region is the area that satisfies all the given constraints. In our exercise, the constraints are: \(x \geq 0\), \(y \geq 0\), and \(x + 2y \leq 5\). To picture this, imagine you're in a two-dimensional space where these constraints serve as boundaries.

• \(x \geq 0\) means we're only considering points right of the y-axis.
• \(y \geq 0\) means points are above the x-axis.
• \(x + 2y \leq 5\) is a line that creates a boundary slope for our region.

The intercepts of this boundary line with the x and y axes help in visualizing the region. When \(x = 0\), \(y\) is 2.5; when \(y = 0\), \(x\) is 5. Together, these conditions define a triangle in the first quadrant where the function applies. Points within or on this triangular border are acceptable for analyzing the optimization problem.

Objective Function

An objective function is what you need to minimize or maximize in an optimization problem. Here, we want to find the minimum and maximum values of the function \( h(x, y) = x^2 + y^2 - 4x - 2y + 1 \).
This function might represent anything from cost to distance, considered within the constraints outlined. We’ll evaluate this function at certain key points within the feasible region, including the perimeter and any critical points within the interior, to find those absolute maximum and minimum values.
Understanding the objective function’s role is crucial as it determines what ‘optimization’ means for each specific problem.

Constraint Analysis

Constraint analysis involves examining the limitations imposed by the problem's constraints. These constraints dictate which solutions are viable, thus defining the feasible region.

For the exercise, the constraints are:

• \(x \geq 0\)
• \(y \geq 0\)
• \(x + 2y \leq 5\)

By analyzing these constraints, we can identify the range where the objective function is defined and can carry significance. Interpret these constraints as the fences within which our search for optimal or extreme values can occur.
This step is crucial because ignoring any constraints might lead to identifying a solution that is not feasible within the problem's context.

Partial Derivatives

Partial derivatives help in finding the points where the function may have extreme values, that is, either a maximum or minimum. For our objective function, the partial derivatives are calculated with respect to \(x\) and \(y\).

- The partial derivative \(\frac{\partial h}{\partial x} = 2x - 4\) shows how the function changes as \(x\) varies, holding \(y\) constant.- The partial derivative \(\frac{\partial h}{\partial y} = 2y - 2\) indicates changes concerning \(y\), with \(x\) fixed.
Setting these derivatives to zero finds potential critical points, which are candidates for extremum within the interior of the feasible region. Solving these equations gives us \(x = 2\) and \(y = 1\). This point \((2, 1)\) needs verification against the boundary points to confirm any extreme value. Thus, partial derivatives are a powerful tool for locating critical points in the optimization process.