Problem 15 - APEX Calculus

Question

Find the absolute maximum and minimum of the function subject to the given constraint. \(f(x, y)=x^{2}+y^{2}+y+1,\) constrained to the triangle with vertices (0,1),(-1,-1) and (1,-1)

Step-by-Step Solution

Answer

Maximum: 5 at (0,1). Minimum: 2 at (-1,-1) or (1,-1).

Step 1: Understand the Problem

We are tasked with finding the absolute maximum and minimum values of the function \( f(x, y) = x^2 + y^2 + y + 1 \)subject to a constraint, where the constraint is that \((x, y)\)must lie within the triangle formed by the vertices \((0,1), (-1,-1), (1,-1)\). We will analyze this problem by considering both the interior and the boundary of the triangle.

Step 2: Check Critical Points Inside the Triangle

First, find the critical points by taking the partial derivatives of\(f(x, y)\).Set \(f_x = 2x = 0\)and \(f_y = 2y + 1 = 0\)to find critical points. This means \(x = 0\)and \(y = -\frac{1}{2}\). Check if \((0, -\frac{1}{2})\)lies within the triangle boundaries.

Step 3: Evaluate Function on the Edges of the Triangle

Evaluate \(f(x, y)\)on the edges of the triangle:1. Between \((0,1)\) and \((-1,-1)\), parameterize: \(x = t-1, y = 1-2t\) for \(0 \leq t \leq 1\).2. Between \((0,1)\) and \((1,-1)\), parameterize: \(x = t, y = 1-2t\) for \(0 \leq t \leq 1\).3. Between \((-1,-1)\) and \((1,-1)\), where \(y = -1\).Find the function value for each edge.

Step 4: Analyze and Compare Values

After calculating each evaluation:- Critical point value: \(f(0, -\frac{1}{2}) = \frac{5}{4}\).- Edge one values: \(f(t-1, 1-2t) = 4t^2 - 4t + 2\), calculate extrema.- Edge two values: \(f(t, 1-2t) = 4t^2 - 4t + 2\), calculate extrema.- Edge three values: \(f(x, -1) = x^2 + 2\) for \(-1 \leq x \leq 1\).Find the maximum and minimum among all these calculated values.

Step 5: Conclusion and Verification

The maximum value is found at the vertex, and similarly the minimum value needs confirmation around boundary or calculated interior points. Confirm these after comparing all calculated values along each path.

Key Concepts

Critical Points

To find critical points of a function, we need to identify where its derivative equals zero, indicating potential peaks, troughs, or saddle points. In the context of functions of multiple variables, like our function \( f(x, y) = x^2 + y^2 + y + 1 \), we take partial derivatives with respect to each variable.

• For \(f_x\), we compute \(\frac{\partial f}{\partial x} = 2x\).
• For \(f_y\), we compute \(\frac{\partial f}{\partial y} = 2y + 1\).

Setting these derivatives to zero, we solve the equations:- \(2x = 0\) gives \(x = 0\).- \(2y + 1 = 0\) gives \(y = -\frac{1}{2}\).
Thus, the only critical point is \((0, -\frac{1}{2})\). It is essential to verify if this point lies within our triangular constraints, which it does. This point is only an initial candidate for being an absolute extremum, so we must be thorough in our overall analysis.

Absolute Extrema

To determine absolute extrema, which are the maximum and minimum values of a function under certain constraints, we need to compare values not just at critical points but also along the boundary of the constraint region.
The absolute extrema of \( f(x, y) = x^2 + y^2 + y + 1 \) are found by calculating the value of the function:

• At the critical points inside the triangle, such as \((0, -\frac{1}{2})\).
• Along the edges of the triangle, parameterizing each edge to a single variable expression.
• At the vertices of the triangle itself, which often present potential extrema values.

In the given problem, the triangle vertices are \((0,1)\), \((-1,-1)\), and \((1,-1)\), and the absolute maximum and minimum can occur at these points or along the triangle's edges due to the nature of the constraint. Therefore, thorough checking of these boundary conditions is essential to ascertain the correct values for any extrema.

Parameterization of Curves

When working with constraints, particularly in a bounded region like a triangle, we often use parameterization to express a line or curve as a function of a single variable. This makes evaluating a constrained function far more manageable.
In this problem, consider the edges of the triangle:

• The line from \((0,1)\) to \((-1,-1)\) is parameterized by \(x = t-1\) and \(y = 1-2t\) for \(0 \leq t \leq 1\).
• The line from \((0,1)\) to \((1,-1)\) uses \(x = t\) and \(y = 1-2t\) with \(0 \leq t \leq 1\).
• Finally, the line from \((-1,-1)\) to \((1,-1)\) maintains a constant \(y = -1\) and varies \(x\) across \([-1,1]\).

Parameterizing these edges allows us to express \(f(x, y)\) in terms of a single variable \(t\), making it simpler to calculate extrema values along each edge.

Partial Derivatives

Partial derivatives involve differentiating a multivariable function with respect to one variable at a time, treating the other variables as constants. They are instrumental in finding critical points and understanding the function’s behavior.
For the function \( f(x, y) = x^2 + y^2 + y + 1 \), the partial derivatives are:

• \(f_x = \frac{\partial f}{\partial x} = 2x\), indicating how \(f\) changes as \(x\) changes. Setting \(f_x = 0\) gives potential critical points for \(x\).
• \(f_y = \frac{\partial f}{\partial y} = 2y + 1\), indicating how \(f\) changes as \(y\) changes. Setting \(f_y = 0\) finds potential critical points for \(y\).

These critical points need verification against constraints and thus play a pivotal role in determining locations where absolute extrema might occur within the defined region.