Question

Find the maximum and minimum values of \(z=2 x^{2}+y^{2}-4 x-2 y+5\) (Figure 3) on the set bounded by the closed triangle with vertices \((0,0),(4,0)\), and \((0,1)\).

Short answer

Minimum is 3 at (1,0); Maximum is 21 at (4,0).

Step-by-step solution

Identify the region and vertices

The given function is constrained by the triangular region with vertices at \((0,0)\), \((4,0)\), and \((0,1)\). These points form the boundaries of the feasible region for the function \(z=2x^2+y^2-4x-2y+5\).

Find critical points within the region

To find the critical points of the function, take the partial derivatives with respect to \(x\) and \(y\) and set them to zero. The partial derivatives are \(\frac{\partial z}{\partial x} = 4x - 4\) and \(\frac{\partial z}{\partial y} = 2y - 2\). Setting these equations to zero gives us the critical point \((1,1)\), which is not within the triangular region, so we do not consider it further.

Evaluate function at the vertices

Calculate the function value at each vertex of the triangular region:- At \((0,0)\), \(z = 5\)- At \((4,0)\), \(z = 21\)- At \((0,1)\), \(z = 8\)

Check function on the boundary lines

Next, we check the boundary lines of the triangle since the critical point is not within the region:- Line from \((0,0)\) to \((4,0)\): Here, \(y=0\) and the function reduces to \(z=2x^2-4x+5\). Complete the square to find that minimum is at \((x=1,0)\) giving \(z=3\).- Line from \((0,0)\) to \((0,1)\): Here, \(x=0\) and the function reduces to \(z=y^2-2y+5\). Completing the square gives the minimum at \((0,1)\) with \(z=8\).- Line from \((0,1)\) to \((4,0)\) parameterized by \((x=4t, y=1-t)\) where \(0 \leq t \leq 1\). Substitute and simplify, but it will not yield a value lower than 3 or higher than 21 on inspection.

Determine maximum and minimum values

From the evaluations and checks:- Minimum value: \(z=3\), occurring at \((1,0)\) on the line \((0,0)\) to \((4,0)\).- Maximum value: \(z=21\), occurring at the vertex \((4,0)\).

Key concepts

Critical Points

Critical points in calculus are crucial for finding the maximum or minimum values a function can take. They are the points where the function's rate of change is zero. To find these points, we calculate the partial derivatives of the function with respect to each variable, set them to zero, and solve the resulting system of equations.
For our function, we have the partial derivative with respect to x, \( \frac{\partial z}{\partial x} = 4x - 4 \), and the partial derivative with respect to y, \( \frac{\partial z}{\partial y} = 2y - 2 \).
Solving these equations gives the critical point (1,1). However, for optimization in a bounded region, it's vital to check if critical points lie within the region of interest. If a critical point is outside the region, as in this case, it is ignored for further analysis.

Partial Derivatives

Partial derivatives help us understand how a function changes as we vary just one of the variables, keeping the others constant. They are denoted as \( \frac{\partial}{\partial x} \) or \( \frac{\partial}{\partial y} \), depending on the variable of interest.
In the provided function \( z=2x^2+y^2-4x-2y+5 \),
the partial derivative with respect to x, \( \frac{\partial z}{\partial x} = 4x - 4 \),
indicates how the function changes as x changes and y remains constant. Similarly, \( \frac{\partial z}{\partial y} = 2y - 2 \) shows changes with respect to y.
By setting the derivatives to zero, we identify where these changes stop, aiding us in finding critical points.

Feasible Region

The feasible region defines the area within which we can search for potential solutions for an optimization problem. It's the set of all points that satisfy the problem's constraints.
In this case, the feasible region is a triangle defined by the vertices \((0,0)\), \((4,0)\), and \((0,1)\).
This means any potential maximum or minimum values of the function must be contained within this triangular area.
Knowing the feasible region is crucial, as it limits our domain and helps us focus only on relevant areas for testing potential solutions.

Boundary Evaluation

Boundary evaluation is studying the function's behavior along the edges of the feasible region. It helps identify if a maximum or minimum occurs on the boundary instead of inside the region.
We examine each edge separately to see how the function behaves. For instance:

• On the line from \((0,0)\) to \((4,0)\), we found that the minimum, \( z=3 \), occurs at \((1,0)\).
• Along the line from \((0,0)\) to \((0,1)\), after simplifying, the value doesn't dip below the vertex point \((0,1)\), where \( z=8 \).
• The line from \((0,1)\) to \((4,0)\) doesn't present any lower minimum than \( z=3 \) or higher maximum than \( z=21 \) upon review.

Overall, exploring boundaries ensures we do not overlook any extreme values that may lie along the edges of the region.