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Chapter 10: Problem 3

Find the absolute maxima and minima of the function \(f(x, y)=x^{2}+\) \(x y+y^{2}\) on the unit circle.

Short Answer

Expert verified
The function has an absolute maximum of \(\frac{2\sqrt{2}}{3}\) and an absolute minimum of \(-\frac{2\sqrt{2}}{3}\) on the unit circle.

Step by step solution

01

Defining the function and the constraint

Firstly, define the original function as \(f(x, y)=x^{2}+xy+y^{2}\) and the constraint, the unit circle, as \(g(x, y)=x^{2}+y^{2}-1=0\).
02

Introducing the Lagrange multiplier

The next step is to define a new function \(L(x, y, \lambda)\), also called the Lagrangian, where \(\lambda\) is the Lagrange multiplier. The Lagrangian is defined as: \(L(x, y, \lambda)=f(x, y)-\lambda(g(x, y)) =x^{2}+xy+y^{2}-\lambda(x^{2}+y^{2}-1)\).
03

Calculating the partial derivatives

The next step is to calculate the partial derivatives of \(L(x, y, \lambda)\) with respect to each variable \(x\), \(y\), and \(\lambda\) and set each equal to zero. These equations result from the fact that at the maximum or minimum point the rate of change will be zero. The equations are as follows: \( \frac{\partial L}{\partial x} = 2x+y-2\lambda x = 0 \) \( \frac{\partial L}{\partial y} = x+2y-2\lambda y = 0 \) \( \frac{\partial L}{\partial \lambda} = 1-x^{2}-y^{2} = 0\).
04

Solve the system of equations

Next, solve the system of simultaneous equations calculated in the previous step. Equation \( \frac{\partial L}{\partial \lambda} = 0 \) will give the unit circle equation \(x^{2}+y^{2} = 1\). Plugging these into equations \( \frac{\partial L}{\partial x} = 0 \) and \( \frac{\partial L}{\partial y} = 0 \) will yield two possible solutions, namely (\(-\frac{\sqrt{6}}{3}, \frac{\sqrt{2}}{3}\)) and ((\(\frac{\sqrt{6}}{3}, -\frac{\sqrt{2}}{3}\)). The corresponding lambda values will be \(\lambda_{1,2} = \pm \frac{\sqrt{3}}{3}\).
05

Find the maximum and minimum values

Substitute these points back into the original function \(f(x, y)\) to determine the absolute maxima and minima. The function's values at the two points are \(f(-\frac{\sqrt{6}}{3}, \frac{\sqrt{2}}{3}) = -\frac{2\sqrt{2}}{3}\) and \(f(\frac{\sqrt{6}}{3}, -\frac{\sqrt{2}}{3}) = \frac{2\sqrt{2}}{3}\). Therefore, the maximum value of the function on the unit circle is \(\frac{2\sqrt{2}}{3}\) and the minimum value is \(-\frac{2\sqrt{2}}{3}\).

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Most popular questions from this chapter

For each problem, locate the critical points and evaluate the Hessian matrix at each critical point. a. \(f(x, y)=(x+y)^{2}\) b. \(f(x, y)=x^{2} y+x y^{2}\) c. \(f(x, y)=x^{4} y+x y^{4}-x y\) d. \(f(x, y, z)=x y+x z+y z\). e. \(f(x, y, z)=x^{2}+y^{2}+x z+2 z^{2}\)

A light ray travels from point A in a medium with index of refraction \(n_{1}\) toward point \(\mathrm{B}\) in a medium with index of refraction \(n_{2}\). Assume that the a. Write the time functional in terms of the travel path \(y(x)\). b. Apply Fermat's Principle of least time to write the Euler Equation for this functional. c. Solve the equation in part b. and show that \(n \sin \theta\) is a constant. d. Let the point at which the light is incident to the interface be at \((x, 0)\). Write an expression for the total time to travel from point \(A\) at \(\left(x_{1}, y_{1}\right)\) to point \(B\) at \(\left(x_{2}, y_{2}\right)\) in terms of the indices of refraction. and the coordinates \(x, x_{1}, x_{2}\), and \(y_{2}\). e. Treating the time as a function of \(x\), minimize this function as a function of one variable and derive Snell's law of refraction.

Given the cylinder defined by \(x^{2}+y^{2}=4\), find the path of shortest length connecting the given points. a. \((2,0,0)\) and \((0,2,5)\) b. \((2,0,0)\) and \((2,0,5)\)

Find the extrema of the given function subject to the given constraint. a. \(f(x, y)=(x+y)^{2}, x^{2}+y=1\) b. \(f(x, y)=x^{2} y+x y^{2}, x^{2}+y^{2}=2\) c. \(f(x, y)=2 x+3 y, 3 x^{2}+2 y^{2}=3\) d. \(f(x, y, z)=x^{2}+y^{2}+z^{2}, x y z=1\) e. \(f(x, y, z)=x y+y x, x^{2}+y^{2}=1, x z=1\)

Use a Lagrange multiplier to find the curve \(x(y)\) of length \(L=\pi\) on the interval \([0,1]\) which maximizes the integral \(I=\int_{0}^{1} y(x) d x\) and pass through the points \((0,0)\) and \((1,0)\).
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