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Chapter 13: Problem 11

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

Short Answer

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

Step by step solution

01

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.
02

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 \((0, -1)\).
03

Evaluate Function at the Critical Point

Substitute \((0, -1)\) into the function: \(f(0, -1) = 2(0)^2 + (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4\).
04

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: \((4x, 2y+2) = \lambda (2x, 2y)\).
05

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.
06

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. \(f(0, -2) = 2(0) + (-2)^2 + 2(-2) - 3 = 1\).
07

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攁reas 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.

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