Calculus-Sec-13-9-Solution

13.9 Lagrange Multiplier by SGLee - WJLee -SWSun

(Ref.http://math1.skku.ac.kr/home/pub/1542/)

Two Examples by http://sfa.sagenb.org/home/pub/133/

Supported by the National Science Foundation under grant DUE-1022574.

Suppose that we wish to ﬁnd the point on the plane that is closest to the point P=(1,2,3). This is the same as ﬁnding the point (x,y,z) that will minimize the distance

f(x,y,z)= ((x−1)2+(y−2)2+(z−3)2)^(1/2) to the point P and satisfy the equation of the plane g(x,y,z)=2x+y−z−5=0.

To solve this problem using Sage, we ﬁrst deﬁne the functions, including h=f−λg.

As a second example, we will use Sage and the method of Lagrange multipliers to ﬁnd the extreme points of f(x, y) = x^2 +3xy + y^2 on the unit circle g(x, y) = x^2 + y^2 -1 = 0.

http://sfa.sagenb.org/home/pub/133/

Notice this time that we have four solutions. To ﬁnd out which of these solutions are maxima and which are minima, we will need to evaluate f(x, y) for each of them.

We can now see that f has two minimum of −1∕2 and two maximum values of 5∕2 on the unit circle. We can summarize our ﬁndings in the following table.

Introduction

One of the most common problems in calculus is that of finding maxima or minima (in general, "extrema") of a function, but it is often difficult to find a closed form for the function being extremized. Such difficulties often arise when one wishes to maximize or minimize a function subject to fixed outside conditions or constraints. The method of Lagrange multipliers is a powerful tool for solving this class of problems without the need to explicitly solve the conditions and use them to eliminate extra variables.

Consider the two-dimensional problem introduced above:

maximize $f(x, y) \,$

subject to $g(x, y) = c. \,$

We can visualize contours of f given by

$f(x, y)=d \,$

for various values of $d$ , and the contour of $g$ given by $g ( x, y ) = c$ .

Suppose we walk along the contour line with $g = c$ . In general the contour lines of $f$ and $g$ may be distinct, so following the contour line for $g = c$ one could intersect with or cross the contour lines of $f$ . This is equivalent to saying that while moving along the contour line for $g = c$ the value of $f$ can vary. Only when the contour line for $g = c$ meets contour lines of $f$ tangentially, do we not increase or decrease the value of $f$ — that is, when the contour lines touch but do not cross.

The contour lines of f and g touch when the tangent vectors of the contour lines are parallel. Since the gradient of a function is perpendicular to the contour lines, this is the same as saying that the gradients of f and g are parallel. Thus we want points $(x,y)$ where $g(x,y) = c$ and

$\nabla_{x,y} f = - \lambda \nabla_{x,y} g$ ,

where

$\nabla_{x,y} f= \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

and

$\nabla_{x,y} g= \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right)$

are the respective gradients. The constant $\lambda$ is required because although the two gradient vectors are parallel, the magnitudes of the gradient vectors are generally not equal.

To incorporate these conditions into one equation, we introduce an auxiliary function

$\Lambda(x,y,\lambda) = f(x,y) + \lambda \cdot \Big(g(x,y)-c\Big),$

and solve

$\nabla_{x,y,\lambda} \Lambda(x , y, \lambda)=0.$

This is the method of Lagrange multipliers. Note that $\nabla_{\lambda} \Lambda(x , y, \lambda)=0$ implies $g(x,y)=c$ .

The constrained extrema of $f$ are critical points of the Lagrangian $\Lambda$ , but they are not local extrema of $\Lambda$ (see Example 2 below).

One may reformulate the Lagrangian as a Hamiltonian, in which case the solutions are local minima for the Hamiltonian. This is done in optimal control theory, in the form of Pontryagin's minimum principle.

The fact that solutions of the Lagrangian are not necessarily extrema also poses difficulties for numerical optimization. This can be addressed by computing the magnitude of the gradient, as the zeros of the magnitude are necessarily local minima, as illustrated in the numerical optimization example.

(Ref. http://en.wikipedia.org/wiki/Lagrange_multiplier 2013.12.11)

1. In each of the following problems, find the extreme(maximum and minimum) values of subjected to the given constrained:

(i) subject to .

(ii) subject to .

(iii) subject to .

(iv) subject to .

(v) subjected to

(vi) subjected to .

(vii) subjected to .

Can you generalize this?

(i)와 (vii)만 풀어보기로 한다.

(i) subject to .

우선 $g(x,y)$ 를 정의하고, $f$ 와 $g$ 의 gradient를 구한다.

$f(x,y)=k g(x,y)$ 와 $g(x,y)=1$ 을 만족하는 $k$ 및 $k$ 에 해당하는 $(x,y)$ 를 구한다.

이 $(x,y)$ 가 critical point이고, 함수 $f(x,y)$ 에 대하여 주어진 범위 내의 최댓값 또는 최솟값이다.

각각의 $(x,y)$ 를 대입했을 때, 다음과 같은 결과가 나온다.

즉, 최댓값은 $\sqrt{13}$ 이고, 최솟값은 $-\sqrt{13}$ 이다.

(vii) subjected to .

위와 같은 방법으로 풀어본다. 변수만 3개로 늘어났을 뿐이다.

즉, 최댓값은 $\sqrt{3}$ 이고, 최솟값은 $-\sqrt{3}$ 이다.

2. Find the points on the sphere that are closest and farthest from the point .

원점을 지나고 방향벡터가 $(1,2,2)$ 인 직선과 $x^2+y^2+z^2=1$ 가 만나는 두 점이

가장 가까운 점 또는 가장 먼 점이 될 것이다.

원점을 지나고 방향벡터가 $(1,2,2)$ 인 직선은 $x=\frac{y}{2}=\frac{z}{2}$ 이다.

즉, $x=t, y=2t, z=2t$ 이다.

즉, $t=\frac{1}{3}$ 일 때, $(\frac{1}{3},\frac{2}{3},\frac{2}{3})$ 로 가장 가까운 점이고,

$t=-\frac{1}{3}$ 일 때, $(-\frac{1}{3},-\frac{2}{3},-\frac{2}{3})$ 로 가장 먼 점이다.

3. Determine the dimensions of a rectangular box, open at the top having volume pf 32 cubic feets and requiring the least amount

of material for its construction.

4. Suppose the cost of manufacturing a particular type of box is such that the base of the box costs three times as much per square

foot as the sides and top. Find the dimensions of the box that minimize the cost for a given volume.

5. Find the maximum and minimum of on the ellipse given by the intersection of the cylinder

and the plane .

NOTE : 제한조건이 두 개인 경우는 $g,h$ 를 잡고,

$\bigtriangledown f=\lambda \bigtriangledown g+\mu \bigtriangledown h$ 를 풀면 된다.

위의 값 중 첫 세 개에는 한 변의 길이가 0인 삼각형이므로 제외해야한다.

즉, $x=y=z=\frac{1}{3}p$ 만이 실질적인 critical point이다.

아래 방정식이 Sage에서 풀리지 않아 손으로 직접구해본 결과

$s=1, t=\pm \frac{1}{2}$ 가 나왔다.

즉, $(-1,1,0)$ 와 $(1,-1,1)$ 가 critical point이다.

$(-1,1,0)$ 일 때, 최댓값을 갖고 $(1,-1,1)$ 일 때, 최솟값을 갖는다.

6. The cone is cut by the plane in some curve . Find the point on that is closest to the origin.

7. In the following exercises find the extreme values of subjected to the two constraints.

(a) ; ; .

(b) ; ; .

(d) ; ; .

8. (a )Use Lagrange multipliers to find the highest and lowest points on the ellipse .

(b) Use Lagrange multipliers to show that the rectangle with maximum area that has a given perimeter is a square.

(d) Find the maximum and minimum volumes of a rectangular box whose surface area is $1200 cm^2$ and whose total

edge length is $.

(e) Use Sage to ind the maximum of , subject to the constraint

(a )Use Lagrange multipliers to find the highest and lowest points on the ellipse .

$f(x,y)=y$ 를 이용한다.

*은 타원이므로 최댓값은 $b$ 이고, 최솟값은 $-b$ 이다.

이와 같이 Lagrange Multiplier로 최댓값과 최솟값을 구하고 있고, 이는 우리가 알고있던 것과 같다.

(b) Use Lagrange multipliers to show that the rectangle with maximum area that has a given perimeter is a square.

우선, 직사각형의 가로의 길이를 $x$ , 세로의길이를 $y$ 라 하고 식을 세워본다.

그러면, $f(x,y)=xy$ 의 $2x+2y=p$ 범위에서 최댓값을 구하는 문제라 할 수 있다.

이 역시 우리가 알고있는 값과 같다.

각 변의 길이가 각각 $a,b,c$ 인 삼각형의 넓이는

$T=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$ 이다.

(Ref. http://en.wikipedia.org/wiki/Heron%27s_formula 2013.12.11)

즉, $f(x,y,z)=\frac{1}{4}\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$ 의 $x+y+z=p$ 에서 최댓값을 가질 조건이

$x=y=z$ 임을 보이면 된다.

9. Find the absolute maximum value and minimum value of the function on the ellipse .

Using ,

Substituting in , then =>

Substituting in => => =>

The absolute maximum value is

The absolute minimum value is

10. Find the absolute maximum value of the function on the ellipse .

Using ,

Substituting in => >

Substituting in => =>

Hence and

Now the absolute maximum value of the function is

수학문화원 설치물

Back to Part II