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Extrema on a circle Find the maximum and minimum values of \(x^{2}+y^{2}\) subject to the constraint \(x^{2}-2 x+y^{2}-4 y=0\)

Short Answer

Expert verified
The extrema values of \(x^2 + y^2\) are both 11.

Step by step solution

01

Interpret the Constraint

The given constraint is an equation for a circle in the form \(x^2 - 2x + y^2 - 4y = 0\). To identify the circle, complete the square for both \(x\) and \(y\).
02

Complete the Square

Rewrite the equation by completing the square:For \(x\):- \(x^2 - 2x = (x-1)^2 - 1\)For \(y\):- \(y^2 - 4y = (y-2)^2 - 4\)The equation becomes \((x-1)^2 - 1 + (y-2)^2 - 4 = 0\).Simplify it to \((x-1)^2 + (y-2)^2 = 5\). This shows the circle is centered at \((1, 2)\) with a radius of \(\sqrt{5}\).
03

Set Up the Objective Function

The objective function is \(x^2 + y^2\). We need to find its maximum and minimum values subject to the constraint \((x-1)^2 + (y-2)^2 = 5\).
04

Utilize Lagrange Multipliers

Using the method of Lagrange multipliers, set up the equations:- \(abla f = \lambda abla g\)- \(f(x, y) = x^2 + y^2\)- \(g(x, y) = (x-1)^2 + (y-2)^2 - 5 = 0\)Find \(abla f = (2x, 2y)\) and \(abla g = (2(x-1), 2(y-2))\).
05

Solve the System of Equations

Equate the gradients:- \(2x = \lambda \, 2(x-1)\)- \(2y = \lambda \, 2(y-2)\)Consider both equations separately:1. \(2x = \lambda \, 2(x-1)\) implies \( x(1 - \lambda) = \lambda \)2. \(2y = \lambda \, 2(y-2)\) implies \( y(1 - \lambda) = 2\lambda \)Solve for \(x\) and \(y\) along with the circle's constraint.
06

Determine Critical Points

By solving the system of equations, find potential critical points by looking at both cases where \(\lambda = 0\) and \(\lambda eq 0\). Integrate with the constraint equation to obtain points \((1, 2+\sqrt{5})\), \((1, 2-\sqrt{5})\), \((1+\sqrt{5}, 2)\), and \((1-\sqrt{5}, 2)\).
07

Evaluate the Objective Function at Each Point

Calculate \(x^2 + y^2\) at each critical point:- For \((1, 2+\sqrt{5})\) and \((1, 2-\sqrt{5})\): \(x^2 + y^2 = 1 + (2\pm\sqrt{5})^2 = 11\)- For \((1+\sqrt{5}, 2)\) and \((1-\sqrt{5}, 2)\): \((1\pm\sqrt{5})^2 + 4 = 11\)All points give \(x^2 + y^2 = 11\).
08

Conclusion

Since every evaluated point results in \(x^2 + y^2 = 11\), the maximum and minimum values of \(x^2 + y^2\) on the given circle are both 11.

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Key Concepts

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

Extrema on a Circle
Imagine you have a circular track and you want to identify the highest and lowest points on this track in terms of their distance from an origin point or another reference point. This is what finding the "extrema on a circle" is all about. These points are where a function reaches its maximum or minimum value, but only along the circle's path.

In the original problem, we are tasked with finding extrema for the function \(x^2 + y^2\) subject to the constraint \(x^2 - 2x + y^2 - 4y = 0\), which represents a circle. Through constraint optimization techniques like Lagrange multipliers, we can find points that offer the maximum and minimum values of the function, but only as long as the points lie on the circle.

It's important to note that these extrema might not be the absolute highest or lowest values of the function over all space—just the highest or lowest ones on this particular path or shape, in this case, the circle.
Completing the Square
Completing the square is a mathematical technique used to transform a quadratic equation into a more recognizable form. This makes it easier to determine the characteristics of shapes such as circles and parabolas.

In the original exercise, the equation \(x^2 - 2x + y^2 - 4y = 0\) is rewritten using completing the square, which simplifies the equation to \((x-1)^2 + (y-2)^2 = 5\).

Through this, it becomes clear that we are working with a circle centered at \((1, 2)\) with a radius of \(\sqrt{5}\). This is because the general form of a circle's equation for a center \((h, k)\) and radius \(r\) is \((x-h)^2 + (y-k)^2 = r^2\). Completing the square helps by simplifying how we interpret the equation and identify the shape.
Constraint Optimization
Constraint optimization involves finding the optimal points (maximum or minimum) for a function but under specific restrictions, known as constraints. In this problem, the constraint defines a circle, and we want to determine where the function \(x^2 + y^2\) reaches its extreme values while staying on the circle.

The method of Lagrange multipliers is a powerful tool used in constraint optimization. It helps to locate points where the objective function's rate of increase is proportional to the rate of increase of the constraint, making sure the optimal or critical points adhere to the given constraints.

Using Lagrange multipliers, you set up a system of equations where the gradient of the function \(f(x, y) = x^2 + y^2\) is equal to a constant, \(\lambda\), times the gradient of the constraint \(g(x, y) = 0\). Solving these equations, you find potential critical points that are evaluated to confirm whether they represent maximum or minimum values.

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Most popular questions from this chapter

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