Problem 11 - Calculus with Concepts in Calculus

Question

Find the extreme values of \(f\) in the region described by the given inequalities. In each case assume that the extreme values exist. $$ f(x, y)=2 x^{2}+y^{2}+2 y-3 ; x^{2}+y^{2} \leq 4 $$

Step-by-Step Solution

Answer

Max: 5 at \((2, 0)\), Min: -4 at \((0, -1)\).

Step 1: Understand the Constraints

The region is defined by the inequality \(x^2 + y^2 \leq 4\). This represents a circle centered at the origin with radius 2. We are tasked with finding the extreme values of the function within and possibly on the boundary of this circle.

Step 2: Finding Critical Points Inside the Region

To find critical points inside the region, we compute the partial derivatives of \(f\). The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 4x\). The partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = 2y + 2\). Setting these equal to zero gives the equations \(4x = 0\) and \(2y + 2 = 0\). Solving these, we find \(x = 0\) and \(y = -1\). This yields a critical point at \((0, -1)\).

Step 3: Evaluate Function at the Critical Point

Substitute \((0, -1)\) into the function: \(f(0, -1) = 2(0)^2 + (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4\).

Step 4: Evaluate Function on the Boundary

On the boundary \(x^2 + y^2 = 4\), we use the method of Lagrange multipliers. Define \(g(x, y) = x^2 + y^2 - 4 = 0\). Set \(abla f = \lambda abla g\). This gives the system: \((4x, 2y+2) = \lambda (2x, 2y)\).

Step 5: Solve the System for Boundary

We have two equations: \(4x = \lambda 2x\) and \(2y + 2 = \lambda 2y\) with \(x^2 + y^2 = 4\). If \(x eq 0\), \(\lambda = 2\). If \(y eq 0\), \(\lambda = \frac{2y + 2}{2y}\). Solving these equations gives possible values: \((\pm 2, 0)\), \((0, \pm 2)\), and \((0, -1)\) on the circle.

Step 6: Evaluate the Function at the Boundary Points

Compute \(f\) at these points:1. \(f(2, 0) = 2(2^2) + 0^2 + 2(0) - 3 = 5\).2. \(f(-2, 0) = 2(-2)^2 + 0 - 3 = 5\).3. \(f(0, 2) = 2(0) + 2^2 + 2(2) - 3 = 5\).4. \(f(0, -2) = 2(0) + (-2)^2 + 2(-2) - 3 = 1\).

Step 7: Identify the Extreme Values

Compare values from the critical point and boundary:- Inside: \(f(0, -1) = -4\).- On boundary: largest value is 5 and smallest is 1.The maximum value of \(f(x, y)\) in the region is 5, and the minimum value is -4.

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes when we vary each variable independently. For the function \(f(x, y) = 2x^2 + y^2 + 2y - 3\), we need to find the partial derivatives to identify critical points. These are points where the function's rate of change is zero in all directions.

• For \(x\), the partial derivative is \(\frac{\partial f}{\partial x} = 4x\). This tells us how \(f\) changes as \(x\) changes while keeping \(y\) constant.


• For \(y\), the partial derivative is \(\frac{\partial f}{\partial y} = 2y + 2\). This tells us how \(f\) changes as \(y\) changes while keeping \(x\) constant.

Setting these derivatives to zero, \(4x = 0\) and \(2y + 2 = 0\), helps us locate critical points—areas where the function might reach a local minimum or maximum.

Lagrange Multipliers

The method of Lagrange multipliers is a tool used to find the extreme values of a function subject to a constraint. For our problem, we want to evaluate \(f(x, y)\) on the circle boundary defined by \(x^2 + y^2 = 4\).

• Define the constraint function as \(g(x, y) = x^2 + y^2 - 4 = 0\).


• Set the gradients \(abla f = \lambda abla g\), where \(\lambda\) is a multiplier that balances both gradients.

This approach leads to a system of equations:

• \(4x = \lambda 2x\)


• \(2y + 2 = \lambda 2y\)


• \(x^2 + y^2 = 4\)

Solving this system gives us potential points on the boundary where extreme values might occur.

Critical Points

Critical points occur where the partial derivatives of a function equal zero. In our exercise, these points are essential to finding local maxima or minima.

• Determine where both partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) are zero.


• For \( f(x, y) = 2x^2 + y^2 + 2y - 3 \), \(4x=0\) gives \(x=0\), and \(2y+2=0\) gives \(y=-1\).

Thus, we have a critical point at \((0, -1)\). Evaluating the function at this point, \(f(0, -1) = -4\), suggests a potential minimum. We must compare this with boundary points to identify absolute extrema.

Boundary Evaluation

When considering potential extreme values of a function, evaluating the boundary of the given constraint is crucial.

For the exercise at hand, the boundary is the circle \(x^2 + y^2 = 4\). We determine extreme values by evaluating \(f(x, y)\) at points obtained by solving with Lagrange multipliers or direct substitution:

• On substituting boundary values, the points of interest are \((\pm 2, 0)\), \((0, \pm 2)\), and previously found critical points.


• Calculate \(f(x, y)\) at each boundary point: \((2,0)\), \((-2,0)\), \((0,2)\), and \((0,-2)\).

This evaluation allows us to calculate values such as \(f(2,0) = 5\), \(f(-2,0) = 5\), and so forth. The comparison of these evaluations with the function value at the internal critical point yields the maximum and minimum values over the defined region.