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Find the absolute maximum and minimum values of the following functions on the given region \(R\). $$f(x, y)=x^{2}+y^{2}-2 y+1 ; R=\left\\{(x, y): x^{2}+y^{2} \leq 4\right\\}$$

Short Answer

Expert verified
Answer: The absolute maximum value of the function is \(9\) at the point \((-2, -1)\), and the absolute minimum value is \(\frac{1}{4}\) at the point \(\left(\frac{1}{2},\frac{3}{2}\right)\).

Step by step solution

01

Find the gradient of both functions

To find the gradient of the given functions, we differentiate each function with respect to \(x\) and \(y\). $$ \nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (2x, 2y - 2) $$ $$ \nabla g = \left(\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right) = (2x, 2y) $$
02

Set the relationship for the Lagrange multiplier

Using the Lagrange multiplier method, we relate the gradients of f and g as follows: $$ \nabla f = \lambda \nabla g $$ Plugging in the gradients calculated in previous step, we get: $$ (2x, 2y - 2) = \lambda (2x, 2y) $$
03

Solve the resulting system of equations for \(x\), \(y\), and \(\lambda\)

From the previous step, we have: $$ 2x = 2x\lambda \hspace{10pt} \text{and} \hspace{10pt} 2y - 2 = 2y\lambda $$ Solving both equations for \(\lambda\) gives: $$ \lambda = \frac{1}{x} \hspace{10pt} \text{and} \hspace{10pt} \lambda = \frac{1}{y - 1} $$ Since both expressions for \(\lambda\) are equal, we can express \(x\) in terms of \(y\): $$ \frac{1}{x} = \frac{1}{y - 1} \implies x = y - 1 $$ Additionally, we know the constraint condition which is \(g(x, y) = 0\): $$ x^2 + y^2 = 4 \implies (y-1)^2 + y^2 = 4 $$ Now solve for \(y\): $$ 2y^2 - 2y - 3 = 0 \implies y = -1, \frac{3}{2} $$ Correspondingly, we find their respective \(x\) values using \(x = y - 1\): $$ x_1 = -2, \hspace{10pt} x_2 = \frac{1}{2} $$ So, we have two points: \((-2, -1)\) and \((\frac{1}{2}, \frac{3}{2})\).
04

Substitute these values back into the function f to find the maximum and minimum values

Now we will evaluate the function \(f(x,y)\) at these two points to find the maximum and minimum values: $$ f(-2, -1) = (-2)^2 + (-1)^2 - 2 (-1) + 1 = 9 $$ and $$ f\left(\frac{1}{2}, \frac{3}{2}\right) = \left(\frac{1}{2}\right)^2 + \left(\frac{3}{2}\right)^2 - 2 \left(\frac{3}{2}\right) + 1 = \frac{1}{4} $$ Comparing these values, we find that the maximum value of the function is \(9\) at point \((-2, -1)\), and the minimum value is \(\frac{1}{4}\) at point \(\left(\frac{1}{2},\frac{3}{2}\right)\).

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Key Concepts

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

Gradient
In optimization problems involving multiple variables, the gradient plays a crucial role. The gradient of a function is essentially a vector that points in the direction of the greatest rate of increase of the function. For the function \(f(x, y) = x^2 + y^2 - 2y + 1\), the gradient \(abla f\) is found by taking the partial derivatives with respect to each variable: \(\frac{\partial f}{\partial x} = 2x\) and \(\frac{\partial f}{\partial y} = 2y - 2\). Hence, the gradient is \((2x, 2y-2)\). \
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In our case, the gradient helps in identifying the "direction" that our function \(f\) changes most rapidly. Similarly, for the constraint \(g(x, y) = x^2 + y^2 - 4\), we find the gradient \(abla g\), which is \((2x, 2y)\). To apply the Lagrange multiplier method, we set these gradients in relation to each other.
Absolute Maximum and Minimum Values
When solving for the absolute maximum and minimum values of a function, we focus on finding the highest and lowest outputs a function can achieve in a given domain. Once we know the potential points of interest, we evaluate our function \(f(x, y) = x^2 + y^2 - 2y + 1\) at these points. \
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In our example, after solving the system of equations derived from the gradients, we discovered two significant points: \((-2, -1)\) and \((\frac{1}{2}, \frac{3}{2})\). By substituting these into the function: \(f(-2, -1) = 9\) gives us the maximum value, whereas \(f(\frac{1}{2}, \frac{3}{2}) = \frac{1}{4}\) gives us the minimum value. \
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This step is important because it confirms whether these points lie on the boundary or inside the region \(R\). Evaluating these ensures we have identified all possibilities for maxima and minima in the constrained region.
Constrained Optimization
Constrained optimization is the process of optimizing an objective function subject to constraints. In our example, we optimized \(f(x, y) = x^2 + y^2 - 2y + 1\) given the constraint \(x^2 + y^2 \leq 4\). This method is ideal when we need to optimize a function within a specific, limited area.\
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To handle constraints, we use a powerful approach called Lagrange multipliers. This technique introduces an additional variable, \(\lambda\), to help us set up a system where the gradient of \(f\) is proportional to the gradient of the constraint function \(g\). The condition \(abla f = \lambda abla g\) ensures that any changes to \(f\) do not violate the constraint. \
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By solving the resulting system of equations, we determined the potential points of optimization. Applied correctly, Lagrange multipliers can simplify seemingly complex problems and find solutions that satisfy all restrictions.

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