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Lecture Comments (5)

0 answers

Post by Joel Fredin on March 2, 2014

I want to find out if it was an ellipse or a hypurbola, but how do you spell that word, you gave a hint. you said "*something* coordinates".

Would be really glad if you replied to this post :)

1 answer

Last reply by: Professor Hovasapian
Thu Aug 1, 2013 1:27 AM

Post by michael Boocher on July 31, 2013

Wow... 6:30 you really should open this section with that point. Total clarity, I was lost until now. Now I don't even need the equations, can logically deduce the requisite operations.

Thnx

1 answer

Last reply by: Professor Hovasapian
Sat Sep 1, 2012 4:23 PM

Post by Mohammed Alhumaidi on September 1, 2012

Can we setup a matrix here to solve all these 5 equations ? To make it simpler ?

Lagrange Multipliers, Continued

Find the maximum and minimum of f(x,y) = 2x − 5y subject to g:x2 + y2 = 1 and h:x − y = 0.
  • To find the maximum and minimum for a multivarible function f subject to two functions g and h we apply the Lagrange multipliers
    ∇f = l1∇g + l2∇h
    g(x,y) = 0
    h(x,y) = 0
  • Now ∇f = (2, − 5), ∇g = (2x,2y) and ∇h = (1, − 1) so that our system of equations is
    (2, − 5) = l1(2x,2y) + l2(1, − 1)
    x2 + y2 − 1 = 0
    x − y = 0
  • Note that x0 or y0 since g and h will not be satisfied, nor if x = y = 0.
  • We can obtain x and y by solving
    x2 + y2 − 1 = 0
    x − y = 0
    as the bottom equation gives x = y, substituting into the top equation results in x2 + ( x )2 − 1 = 0.
  • Hence x = ±[1/(√2 )] = y. Our two value points are f( [1/(√2 )],[1/(√2 )] ) = − [3/(√2 )] and f( − [1/(√2 )], − [1/(√2 )] ) = [3/(√2 )].
  • Should we check that l1 and l2 are defined? No, this is not necessary as g and h are a circle and line that intersect only at two points. As such f obtains extrema only at two points.
  • These two points must be ( [1/(√2 )],[1/(√2 )] ) and ( − [1/(√2 )], − [1/(√2 )] ) and so we can confidently state that l1 and l2 are defined (because the points exist). This will not be the case with intersecting surfaces.
The maximum value is [3/(√2 )] while the minimum value is − [3/(√2 )].
Find the maximum and minimum of f(x,y) = 3y subject to g:x2 + y2 = 1 and h:2x + y = 0.
  • To find the maximum and minimum for a multivarible function f subject to two functions g and h we apply the Lagrange multipliers
    ∇f = l1∇g + l2∇h
    g(x,y) = 0
    h(x,y) = 0
  • Now ∇f = (0,3), ∇g = (2x,2y) and ∇h = (2,1) so that our system of equations is
    (0,3) = l1(2x,2y) + l2(2,1)
    x2 + y2 − 1 = 0
    2x + y = 0
  • Note that x0 or y0 since g and h will not be satisfied, nor if x = y = 0.
  • We can obtain x and y by solving
    x2 + y2 − 1 = 0
    2x + y = 0
    as the bottom equation gives y = − 2x, substituting into the top equation results in x2 + ( − 2x )2 − 1 = 0.
  • Hence x = ±[1/(√5 )] with y = − 2( [1/(√5 )] ) = − [2/(√5 )] and y = − 2( − [1/(√5 )] ) = [2/(√5 )]. Our two value points are f( [2/(√5 )], − [2/(√5 )] ) = − [6/(√5 )] and f( − [2/(√5 )],[2/(√5 )] ) = [6/(√5 )].
  • Should we check that l1 and l2 are defined? No, this is not necessary as g and h are a circle and line that intersect only at two points. As such f obtains extrema only at two points.
  • These two points must be ( [2/(√5 )], − [2/(√5 )] ) and ( − [2/(√5 )],[2/(√5 )] ) and so we can confidently state that l1 and l2 are defined (because the points exist). This will not be the case with intersecting surfaces
The maximum value is [6/(√5 )] while the minimum value is − [6/(√5 )].
Find the maximum and minimum of f(x,y) = 4x2 − y subject to g:x − 3y = 0 and h:y2 + 2x2 = 1.
  • To find the maximum and minimum for a multivarible function f subject to two functions g and h we apply the Lagrange multipliers
    ∇f = l1∇g + l2∇h
    g(x,y) = 0
    h(x,y) = 0
  • Now ∇f = (8x, − 1), ∇g = (1, − 3) and ∇h = (4x,2y) so that our system of equations is
    (8x, − 1) = l1(1, − 3) + l2(4x,2y)
    x − 3y = 0
    y2 + 2x2 = 1
  • Note that x0 or y0 since g and h will not be satisfied, nor if x = y = 0.
  • We can obtain x and y by solving
    x − 3y = 0
    y2 + 2x2 = 1
    as the top equation gives x = 3y, substituting into the bottom equation results in y2 + 2( 3y )2 − 1 = 0.
  • Hence y = ±[1/(√{19} )] with x = 3( [1/(√{19} )] ) = [3/(√{19} )] and y = 3( − [1/(√{19} )] ) = − [3/(√{19} )]. Our two value points are f( [3/(√{19} )],[1/(√{19} )] ) = [36/19] − [1/(√{19} )] and f( − [3/(√{19} )], − [1/(√{19} )] ) = [36/19] + [1/(√{19} )].
  • Should we check that l1 and l2 are defined? No, this is not necessary as g and h are an ellipse and line that intersect only at two points. As such f obtains extrema only at two points.
  • These two points must be ( [3/(√{19} )],[1/(√{19} )] ) and ( − [3/(√{19} )], − [1/(√{19} )] ) and so we can confidently state that l1 and l2 are defined (because the points exist). This will not be the case with intersecting surfaces.
The maximum value is [36/19] + [1/(√{19} )] while the minimum value is [36/19] − [1/(√{19} )].
Find the maximum and minimum of f(x,y,z) = x2 + y2 + z2 subject to g:x + z = 1 and h:x − y = 2.
  • To find the maximum and minimum for a multivarible function f subject to two functions g and h we apply the Lagrange multipliers
    ∇f = l1∇g + l2∇h
    g(x,y,z) = 0
    h(x,y,z) = 0
  • Now ∇f = (2x,2y,2z), ∇g = (1,0,1) and ∇h = (1, − 1,0) so that our system of equations is
    (2x,2y,2z) = l1(1,0,1) + l2(1, − 1,0)
    x + z − 1 = 0
    x − y − 2 = 0
  • Note that if x = 0 then z = 1 and y = − 2 and f(0,1, − 2) = √5 . If x = z = 0 we have − 1 = 0, a contradiction. Similarly if x = y = 0 we obtain − 2 = 0.
  • We now verify if the value point (0,1, − 2) satisfies (
    2x
    2y
    2z
    ) = l1(
    1
    0
    1
    ) + l2(
    1
    − 1
    0
    ), that is (
    0
    2
    − 4
    ) = l1(
    1
    0
    1
    ) + l2(
    1
    − 1
    0
    ).
  • The second component gives l2 = − 2 while the third component gives l1 = − 4. According to the first component 0 = l1 + l2 = − 4 − 2 = − 6 but 0 − 6. Hence our value point (0,1, − 2) is not an extrema.
  • If y = 0 then x = 2 and z = − 1 and f(2,0, − 1) = √5 . If y = z = 0 we have x = 14 and x = 2, a contradiction.
  • We now verify if the value point (2,0, − 1) satisfies (
    2x
    2y
    2z
    ) = l1(
    1
    0
    1
    ) + l2(
    1
    − 1
    0
    ), that is (
    4
    0
    − 2
    ) = l1(
    1
    0
    1
    ) + l2(
    1
    − 1
    0
    ).
  • The second component gives l2 = 0 while the third component gives l1 = − 2. According to the first component 4 = l1 + l2 = − 2 + 0 = − 2 but 4 − 2. Hence our value point (2,0, − 1) is not an extrema.
  • If z = 0 then x = 1 and y = − 1 and f(1, − 1,0) = √2 .
  • We now verify if the value point (1, − 1,0) satisfies (
    2x
    2y
    2z
    ) = l1(
    1
    0
    1
    ) + l2(
    1
    − 1
    0
    ), that is (
    2
    − 2
    0
    ) = l1(
    1
    0
    1
    ) + l2(
    1
    − 1
    0
    ).
  • The second component gives l2 = 2 while the third component gives l1 = 0. According to the first component 2 = l1 + l2 = 0 + 2 = 2 so our value point (1, − 1,0) is an extrema.
  • We now assume that x0, y0 and z0. (2x,2y,2z) = l1(1,0,1) + l2(1, − 1,0) is equivalent to
    2x = l1 + l2
    2y = − l2
    2z = l1
    substituting the second and third equation into the first yields 2x = 2z − 2y or x + y − z = 0.
  • We now have the system
    x + y − z = 0
    x + z − 1 = 0
    x − y − 2 = 0
    whose solution is x = 1, y = − 1 and z = 0. Note that this is the same point we obtained from before.
  • Our only extrema is then f(1, − 1,0) = √2 .
The minimum value is √2 , since (2,0, − 1) = √5 and satisfies both g and h and √5 > √2 .
Use Lagrange multipliers to set up a system of equations \protect
∇f = l1∇g + l2∇h
g(x,y,z) = 0
h(x,y,z) = 0\protect
that finds the maximum and minimum value of f(x,y,z) = xz − yz subject to g:x2 − y − z = 0 and h:y2 + z + x = 0.
  • Computing ∇f = l1∇g + l2∇h gives (z, − z,x − y) = l1(2x, − 1, − 1) + l2(2y,1,1) which is equivalent to
    z = 2xl1 + 2yl2
    − z = − l1 + l2
    x − y = − l1 + l2
Our system of equations is \protect
z = 2xl1 + 2yl2
z = l1 − l2
x − y = − l1 + l2
x2 − y − z = 0
y2 + z + x = 0\protect
Use Lagrange multipliers to set up a system of equations \protect
∇f = l1∇g + l2∇h
g(x,y,z) = 0
h(x,y,z) = 0\protect
that finds the maximum and minimum value of f(x,y,z) = x2 + y2 + z2 subject to g:x2 + y2 − z = 3 and h:y2 + z2 − x = 2.
  • Computing ∇f = l1∇g + l2∇h gives (2x,2y,2z) = l1(2x,2y, − 1) + l2( − 1,2y,2z) which is equivalent to
    2x = 2xl1 − l2
    2y = 2yl1 + 2yl2
    2z = − l1 + 2zl2
Our system of equations is
2x = 2xl1 − l2
y = yl1 + yl2
2z = − l1 + 2zl2
x2 + y2 − z − 3 = 0
y2 + z2 − x − 2 = 0
Use Lagrange multipliers to set up a system of equations \protect
∇f = l1∇g + l2∇h
g(x,y,z) = 0
h(x,y,z) = 0\protect
that finds the maximum and minimum value of f(x,y,z) = [(x2)/9] + y2 − [(z2)/16] subject to g:x − 2y − z2 = 0 and h:x + y + z = − 1.
  • Computing ∇f = l1∇g + l2∇h gives ( [2x/9],2y, − [z/8] ) = l1(1, − 2, − 2z) + l2(1,1,1) which is equivalent to
    [2x/9] = l1 + l2
    2y = − 2l1 + l2
    − [z/8] = − 2zl1 + l2
Our system of equations is
[2x/9] = l1 + l2
2y = − 2l1 + l2
− [z/8] = − 2zl1 + l2
x − 2y − z2 = 0
x + y + z + 1 = 0
Find the maximum and minimum of f(x,y) = xyz subject to g:3x − y = 1 and h:y − 3z = 1.
  • To find the maximum and minimum for a multivarible function f subject to two functions g and h we apply the Lagrange multipliers
    ∇f = l1∇g + l2∇h
    g(x,y,z) = 0
    h(x,y,z) = 0
  • Now ∇f = (yz,xz,xy), ∇g = (3, − 1,0) and ∇h = (0,1, − 3) so that our system of equations is
    (yz,xz,xy) = l1(3, − 1,0) + l2(0,1, − 3)
    3x − y − 1 = 0
    y − 3z − 1 = 0
  • Note that if x = 0 then y = − 1 and z = − [2/3] and f( 0, − 1, − [2/3] ) = 0. If x = z = 0 we have y = − 1 and y = 1, a contradiction. Similarly if x = y = 0 we obtain − 1 = 0.
  • We now verify if the value point ( 0, − 1, − [2/3] ) satisfies (
    yz
    xz
    xy
    ) = l1(
    3
    − 1
    0
    ) + l2(
    0
    1
    − 3
    ), that is (
    2 \mathord/ phantom 2 33
    0
    0
    ) = l1(
    3
    − 1
    0
    ) + l2(
    0
    1
    − 3
    ).
  • The third component gives l2 = 0 and so the second component gives l1 = 0. According to the first component [2/3] = 3l1 = 3(0) = 0 but [2/3]0. Hence our value point ( 0, − 1, − [2/3] ) is not an extrema.
  • If y = 0 then x = [1/3] and z = − [1/3] and f( [1/3],0, − [1/3] ) = 0. If y = z = 0 we have − 1 = 0, a contradiction.
  • We now verify if the value point ( [1/3],0, − [1/3] ) satisfies (
    yz
    xz
    xy
    ) = l1(
    3
    − 1
    0
    ) + l2(
    0
    1
    − 3
    ), that is (
    0
    1 \mathord/ phantom 1 99
    0
    ) = l1(
    3
    − 1
    0
    ) + l2(
    0
    1
    − 3
    ).
  • The third component gives l2 = 0 and the first component gives l1 = 0. According to the second component − [1/9] = − l1 + l2 = − (0) + 0 = 0 but − [1/9]0. Hence our value point ( [1/3],0, − [1/3] ) is not an extrema.
  • If z = 0 then y = 1 and x = [2/3] and f( [2/3],1,0 ) = 0.
  • We now verify if the value point ( [2/3],1,0 ) satisfies (
    yz
    xz
    xy
    ) = l1(
    3
    − 1
    0
    ) + l2(
    0
    1
    − 3
    ), that is (
    0
    0
    2 \mathord/ phantom 2 33
    ) = l1(
    3
    − 1
    0
    ) + l2(
    0
    1
    − 3
    ).
  • The third component gives l2 = − [2/9] and the first component gives l1 = 0. According to the second component 0 = − l1 + l2 = − (0) + ( − [2/9] ) = 0 but 0 − [2/9]. Hence our value point ( [2/3],1,0 ) is not an extrema.
  • We now assume that x0, y0 and z0. Our strategy is to solve for x and y in terms of z.
  • Note that 3x − y − 1 = 0 is equivalent to x = [(y + 1)/3] and y − 3z − 1 = 0 is equivalent to y = 1 + 3z, so x = [(y + 1)/3] = [((1 + 3z) + 1)/3] = z + [2/3].
  • Now (yz,xz,xy) = l1(3, − 1,0) + l2(0,1, − 3) is equivalent to
    yz = 3l1
    xz = − l1 + l2
    xy = − 3l2
    substituting the first and third equation into the second yields xz = − ( [yz/3] ) + ( − [xy/3] ) or 3xz = − yz − xy.
  • Using 3xz = − yz − xy and our equalities x = z + [2/3], y = 1 + 3z we can substitute x and y in order to have a quadratic equation in terms of z, z2 + [2/3]z + [2/27] = 0.
  • The quadratic formula gives us the solutions z = [( − 3 ±√3 )/9].
  • Then x = z + [2/3] = ( [( − 3 + √3 )/9] ) + [2/3] = [(3 + √3 )/9] and y = 1 + 3z = 1 + 3( [( − 3 + √3 )/9] ) = [(√3 )/3] gives f( [(3 + √3 )/9],[(√3 )/3],[( − 3 + √3 )/9] ) = [( − 2√3 )/81].
  • Also x = z + [2/3] = ( [( − 3 − √3 )/9] ) + [2/3] = [(3 − √3 )/9] and y = 1 + 3z = 1 + 3( [( − 3 − √3 )/9] ) = − [(√3 )/3] gives f( [(3 − √3 )/9], − [(√3 )/3],[( − 3 − √3 )/9] ) = [(2√3 )/81].
  • Our possible extrema are f( [(3 + √3 )/9],[(√3 )/3],[( − 3 + √3 )/9] ) = [( − 2√3 )/81] and f( [(3 − √3 )/9], − [(√3 )/3],[( − 3 − √3 )/9] ) = [(2√3 )/81].
The maximum value is [(2√3 )/81] while the minimum value is − [(2√3 )/81].
Find the minimum value of f(x,y,z) = 2x − 5y + z on the curve of intersection of x + y + z = 1 and x2 + y2 = 1.
  • To find the maximum of f subject to two functions g:x + y + z − 1 = 0 and h:x2 + y2 − 1 = 0 we apply the Lagrange multipliers
    ∇f = l1∇g + l2∇h
    g(x,y,z) = 0
    h(x,y,z) = 0
  • Now ∇f = (2, − 5,1), ∇g = (1,1,1) and ∇h = (2x,2y,0) so that our system of equations is
    (2, − 5,1) = l1(1,1,1) + l2(2x,2y,0)
    x + y + z − 1 = 0
    x2 + y2 = 1
    , more precisely
    2 = l1 + 2xl2
    − 5 = l1 + 2yl2
    1 = l1
    x + y + z − 1 = 0
    x2 + y2 = 1
  • Now, since l1 = 1 then 2 = l1 + 2xl2 = 1 + 2xl2 or x = [1/(2l2)]. Also − 5 = l1 + 2yl2 = 1 + 2yl2 or y = − [3/(l2)].
  • Substituting into x2 + y2 = 1 gives ( [1/(2l2)] )2 + ( − [3/(l2)] )2 = 1 that results in l2 = ±[(√{37} )/2].
  • Then x = [1/(2l2)] = [1/2]( [2/(√{37} )] ) = [1/(√{37} )] and y = − [3/(l2)] = − 3( [2/(√{37} )] ) = − [6/(√{37} )]. Using x + y + z − 1 = 0 yields z = [(5 + √{37} )/(√{37} )] thus f( [1/(√{37} )], − [6/(√{37} )],[(5 + √{37} )/(√{37} )] ) = 1 + √{37} .
  • Also x = [1/(2l2)] = [1/2]( − [2/(√{37} )] ) = − [1/(√{37} )] and y = − [3/(l2)] = − 3( − [2/(√{37} )] ) = [6/(√{37} )]. Using x + y + z − 1 = 0 yields z = [( − 5 + √{37} )/(√{37} )] thus f( − [1/(√{37} )],[6/(√{37} )],[( − 5 + √{37} )/(√{37} )] ) = 1 − √{37} .
The minimum value is 1 − √{37} .
Find the minimum value of f(x,y,z) = x2 + y2 + z2 on the curve of intersection of x2 + y2 = z and z = 1.
  • To find the minimum of f subject to two functions g:x2 + y2 − z = 0 and h:z − 1 = 0 we apply the Lagrange multipliers
    ∇f = l1∇g + l2∇h
    g(x,y,z) = 0
    h(x,y,z) = 0
  • Now ∇f = (2x,2y,2z), ∇g = (2x,2y, − 1) and ∇h = (0,0,1) so that our system of equations is
    (2x,2y,2z) = l1(2x,2y, − 1) + l2(0,0,1)
    x2 + y2 − z = 0
    z − 1 = 0
    , more precisely
    2x = 2xl1
    2y = 2yl1
    2z = − l1 + l2
    x2 + y2 − z = 0
    z − 1 = 0
  • We know that z = 1. If x = 0, then y = ±√z = ±1 and obtain the value points (0,1,1) and (0, − 1,1). Note that x = 0 = y yields the contradiction 0 = z = 1. Hence both x and y cannot be zero.
  • Now we check if (0,1,1) satisfies (
    2x
    2y
    2z
    ) = l1(
    2x
    2y
    − 1
    ) + l2(
    0
    0
    1
    ) so (
    0
    2
    2
    ) = l1(
    0
    2
    − 1
    ) + l2(
    0
    0
    1
    ).
  • Our second component gives l1 = 1 and so our third component gives 2 = − 1 + l2 or l2 = 3. Since l1 and l2 are defined, our value point f(0,1,1) = 2 is an extrema.
  • Also, we check if (0, − 1,1) satisfies (
    2x
    2y
    2z
    ) = l1(
    2x
    2y
    − 1
    ) + l2(
    0
    0
    1
    ) so (
    0
    − 2
    2
    ) = l1(
    0
    − 2
    − 1
    ) + l2(
    0
    0
    1
    ).
  • Our second component gives l1 = 1 and so our third component gives 2 = − 1 + l2 or l2 = 3. Since l1 and l2 are defined, our value point f(0, − 1,1) = 2 is an extrema.
  • If y = 0, then x = ±√z = ±1 and obtain the value points (1,0,1) and ( − 1,0,1).
  • Now we check if (1,0,1) satisfies (
    2x
    2y
    2z
    ) = l1(
    2x
    2y
    − 1
    ) + l2(
    0
    0
    1
    ) so (
    2
    0
    2
    ) = l1(
    2
    0
    − 1
    ) + l2(
    0
    0
    1
    ).
  • Our first component gives l1 = 1 and so our third component gives 2 = − 1 + l2 or l2 = 3. Since l1 and l2 are defined, our value point f(1,0,1) = 2 is an extrema.
  • Also, we check if ( − 1,0,1) satisfies (
    2x
    2y
    2z
    ) = l1(
    2x
    2y
    − 1
    ) + l2(
    0
    0
    1
    ) so (
    − 2
    0
    2
    ) = l1(
    − 2
    0
    − 1
    ) + l2(
    0
    0
    1
    ).
  • Our first component gives l1 = 1 and so our third component gives 2 = − 1 + l2 or l2 = 3. Since l1 and l2 are defined, our value point f( − 1,0,1) = 2 is an extrema.
  • It is interesting that four of our value points are extrema and give the value 2. In fact if x0 and y0. We obtain the same results: l1 = 1 and l2 = 3 and all points (x,y,1) satisfying x2 + y2 = 1 as extrema with value 2.
Hence, the circle x2 + y2 = 1 contains all value points that yield the minimum 2.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Lagrange Multipliers, Continued

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Lagrange Multipliers 0:42
    • First Example of Lesson 20
    • Let's Look at This Geometrically
  • Example 1: Lagrange Multiplier Problem with 2 Constraints 8:42
    • Part 1: Question
    • Part 2: What We Have to Solve
    • Part 3: Case 1
    • Part 4: Case 2
    • Part 5: Final Solution

Transcription: Lagrange Multipliers, Continued

Hello and welcome back to educator.com and multi-variable calculus.0000

In today's lesson we are going to continue our discussion of Lagrange multipliers.0004

I am going to start by revisiting a previous example and describing the geometry of the solution.0008

The reason that I want to do that is so that you can see why it is that this criterion of gradient f equals some λ × the gradient of g, why it actually makes sense.0013

I just wanted you to see what the geometry actually looks like, so it does not look like it just fell out of the sky.0025

I encourage you to look at the proofs, or some of the explanations in your textbooks if you want something a little bit more rigorous and a little bit more accurate, but I do want it to make sense geometrically what it is that is happening.0031

So, let us go ahead and get started. Let us see, in the first example of the last lesson, we found... so let me just write all of this out... in the first example of lesson 20, we found the max and the min... you know what, let me not call them max and min, let me call them extrema.0044

We found the extrema values of f(x,y), and if you recall that function was equal to x × y subject to the constraint that the point (x,y) must also satisfy.0080

This function g(x,y), which was x2/8 + y2/2 = 1. We were maximizing or minimizing this function subject to the constraint that the x's and y's that actually go into this that maximize it and minimize it also happen to lie on this ellipse. That is what we were doing... equals 1.0121

Okay. We found four points that satisfied... okay, we had the point (2,1), (2,-1), (-2,1), and (-2,-1).0144

The respective f values, in other words when we put these points into x, y, the respective f values, I am just going to write them underneath... we had 2, -2, -2, and 2.0170

These two were the points where the function achieved a maximum, and these two were the points where it achieved a minimum. 0185

Let us take a look at what this looks like geometrically. So, let us look at this geometrically.0193

What you have -- I am just going to draw a little x, y plane here -- and again we are looking for points (x,y) in the x, y, plane that actually maximize this function x, y but also happen to lie on the ellipse.0209

Well, x, y, if you do the level curves of x, y... in other words x, y, equals some constant, x,y = 3, x, y = 2, x, y = 5/2, whatever. You are going to get a series of hyperbolas. That is what you are going to end up getting.0222

I am going to go ahead and draw a few. First we want to make sure that x and y actually satisfy this equation. In other words, that they lie on this ellipse.0238

Let me just go ahead and draw an ellipse here... I am not very good at drawing at ellipses, but again, the drawing itself is not that important. 0249

So, this is x2 + y2 = 2. Where you have this right here is the distance from the center to this vertex is going to be sqrt(8), and this one is going to be sqrt(2).0258

The points x,y, that lie on here, which ones are the ones that actually maximize? Well, we found these points right here. 0271

(2,1), there was one, (-2,1), (-2,-1), and (2,-1), so here is actually what it looks like geometrically... oops, not very good, let us make it a little bit better than that... here is that... here is another point.0277

So, in other words, if you were to look at this at the point (2,1), (-2,1), (-2,-1), and (2,-1), these are the points that maximize the function. These are also the points that happen to lie on this ellipse, so the constraint is satisfied. Those are the points that we found.0300

Well, geometrically, what it is, take a look at this as it turns out at this point if I were to actually take the tangent line to both curves as it turns out, they correspond. 0323

In other words, if I were to take this tangent line at this point happens to be the same for this curve, g, and for this curve, f.0338

What ends up happening is the following: the gradient, at those points, looks like this. The gradient of f will be this way, the gradient of g will be this way.0351

Because the tangent lines correspond, the gradient vector is perpendicular to the tangent line, so as it turns out they're moving in the same direction.0360

Well, one is going to be either the same length or it is going to be longer than the other. That is what the Lagrange multiplier is. It is just a scalar multiple of the gradient vector.0369

When we say that the gradient of f equals some multiplier λ × the gradient g, what we are saying is they are either moving in the same direction or the opposite direction depending on whether λ is positive or negative.0380

In other words the gradient vector g or the gradient vector of f is a scalar multiple of the gradient vector g.0394

The gradient vector g right here looks something like this. Here, so this was the max and this was a max. Notice the gradient vectors are in the same direction.0402

In the case of the mins, the gradient vectors are actually pointing in opposite directions, but it is still just a scalar multiple of the other.0413

In other words, they have the same direction either this way or this way, but they are part of the same line and they are on the same line because the tangent plane, or the tangent line itself to the curves at that point that satisfies the max and min, they are the same for both curves.0421

So, the gradient are multiples of each other. The gradient vectors are multiples of each other, and that is what is happening. That is what this actually says.0443

I just wanted to demonstrate geometrically what it is that is going on.0452

Again, it is a good idea to see that to get an idea of where it comes from, but again, it is the algebra that matters, because when you move on to higher dimensions, you are not going to quite have the geometric intuition that you can get.0456

For example, this works in 2 dimensions. If we were working with a 3-dimensional example, a function of 3 variables, we would have x, y, and z, and then we would be talking about surfaces.0470

Where you have one surface and then another surface, the point p that happens to be an extremum is going to be the point where those 2 surfaces actually touching, and the tangent plane to those surfaces is going to be the same. 0480

So the gradient vector is going to be perpendicular to that tangent plane and the gradient vector, they're either going to be in the same direction, or they are going to be in opposite directions, but they are going to lie along the same line.0491

That is it, that is all that I wanted to demonstrate. We are going to do one last example for Lagrange multipliers, except this time what we are going to do is we are going to have 2 constraints instead of 1.0502

So, we are going to try to maximize or minimize a function subject not just to 1 constraint, but another. Let us go ahead and do that example. Okay. Example 1.0512

So, let us see. A Lagrange multiplier problem with 2 constraints, so I am just going to describe how it is that we actually deal with it before we start the example.0533

A Lagrange multiplier problem with 2 constraints, we introduce 2 multipliers, that is it. The number of constraints, you just introduce another multiplier and the equation actually ends up being the same, but just with another multiplier.0561

You end up having just another equation to solve. So, introduce 2 multipliers, we will call them λ1 and λ2, so, we must solve the following system.0574

Our problem comes down to solving the gradient of f at... let me actually write it all out here, a little further to the left so that I do not have to squeeze anything in.0603

The gradient of f at the extrema = λ × the gradient of g1 at p, so now we have 2 constraints, g1 and g2 + λ2 × gradient of g2 at p. That is it.0616

We basically just add one more thing. This is the normal one, gradient f = λ × gradient g, but now since we have 2 constraints, g1 and g2, we just add one more term.0635

We also solve g1 of x, y, z. I am just going to call it point g1 of x, y, z, and again I am just using a 3-dimensional example = 0, and g2 of x, y, z = 0.0647

This system right here is the system we end up having to solve. That is it, just one more equation, and this is our one more equation and this is our one extra term.0670

So, now we will go ahead and start the problem.0680

The cone z2 = x2 + y2 is cut by the plane z = 1 + x + y in a conic section.0684

Hopefully you remember from your algebra and pre-calculus days, and perhaps even in single-variable calculus, when you have a cone in 3-space and you slice through that cone, you are going to end up getting the conic section graphs.0709

You are either going to get a circle, you are going to get an ellipse, or you are going to get a hyperbola.0725

Or, in certain, special cases, depending you are going to get a straight line. Okay... conic section, alright.0732

So, we want to find the points on this conic section, and notice we have not said anything about what this conic section is. We just know that it is a conic section.0739

We do not know whether it is a conic section, ellipse, or hyperbola.0754

So, conic section that are nearest and farthest from the origin.0758

We have this cone in space, we slice it with a plane, and there are going to be points that are on this conic section that are going to satisfy both the equation for the cone and the equation for the plane.0776

A couple of those points, or one of those points, however many, some are going to be furthest from the origin, some are going to be closest to the origin, so, that is what we are doing. 0791

Now, we need to find the function that we are going to maximize in this particular case notice this actually ends up saying nothing about.. it set it in a different form.0799

It did not give us the function that we are trying to maximize or minimize explicitly, we have to be able to extract it from the problem itself.0808

Well, since we are finding the nearest and furthest point from the origin, we are going to look to the distance function, but we are not going to deal with the distance function as it is with the radical sign. 0815

We are just going to square that distance function because if you minimize the distance, you are also minimizing the square of that distance, because again, we do not want to deal with radical signs, we want to make the math as tractable as possible.0827

Okay. SO the function that we are going to be dealing with, f itself, is going to be a function of 3 variables, and it is going to be the distance function, so x2 + y2 + z2, 0839

But again, we do not want to deal with the radical, so minimizing or maximizing this is the same as minimizing or maximizing this, just the square of it. x2 + y2 + z2, so that is our f.0853

Now our g1 of x, y, z, that is going to be x2 + y2 - z2... basically what we do is we take this equation and we move everything over to one side and set it equal to 0... - z2 = 0. That is the other equation that we have to satisfy.0863

Then of course we have a second constraint, so we also have to satisfy this equation. g(x,y,z) = 1 + x + y -z = 0. That is of course, this equation, again, with everything moved over to 1 side, set equal to 0. 0884

It always a good idea to write the equation that you are actually dealing with. The f and the particular g's, in this case we have 2 g's that we also need to satisfy.0905

Now, we have to solve the gradient of f = λ1 × the gradient of g1 + λ2 × gradient of g2... and I apologize if I kind of keep writing things over and over again.0916

I think that it is important to do that because often time in mathematics there is so much going on, we try to take shortcuts by not writing everything down and that is a bad idea.0940

Take the couple of extra seconds. If you have to rewrite an equation 2 or 3 times, that is not a problem. It is a good way to solidify it in your mind, to get a good grasp on it. You need to get in the habit of writing things out. The more shortcuts you take, the more chance that you are going to make a mistake.0948

Again, as you have already noticed with the last two lessons, these Lagrange multiplier problems, they are not difficult, there is just a lot going on, on the page, so you want to keep track of everything.0965

Okay. So, we have to solve this, that is the one equation, so we have to solve g1 = 0, and g2 = 0, so let us go ahead and come up with this one here. 0975

When we take the gradient of f, we are going to get 2x, 2y, 2z, and again, just a personal taste for me, I tend to write them in terms of column vectors instead of row vectors -- your choice, whatever works best for you -- equals λ1 × the gradient of g1, which is 2x, 2y, and -2z + λ2 × 1, 1, and -1.0986

These are the 5 equations that we are going to have to solve simultaneously.1018

The first equation, let me do this in blue. We have this equation from the first row, from the second row, and from the third row... so we have... and I am actually going to number these, I am going to refer to them as numbers instead of continuously writing them down.1022

So, equation 1, we have 2x = 2λ1x + λ2.1036

Equation 2 we have 2y = 2λ1y + λ2.1046

Then we have the third equation which is 2z = -2 × λ1z - λ2.1056

Then, of course, we have the g1 and g2, so we have x2 + y2 - z2 = 0, and we have 1 + x + y -z = 0.1064

These are the... this is the system that we have to solve simultaneously, solve simultaneously 1086

So, let us just get started. Again, when you are dealing with this many simultaneous equations, there really is no algorithmic approach to this.1097

You just have to sort of use all of the resources that you have at your disposal, try a bunch of different things... we do not want to give you the impression that just by looking at this I automatically know what to do. 1106

I mean, I have already worked this out. When I look at this on a page, I just automatically know do this, do this, do this, do this... it is not like that. We do not want to give you that impression.1118

It takes time. It takes some time, it takes some experience. I still struggle with these myself, because, again, you have got 5 equations to solve. You have different cases to consider, different things to look at... so by all means do not worry about the time that it takes and do not worry that it is difficult. 1128

On an exam of course, it is not going to be something like this. You are going to have something a little bit simpler to work with. These are just bigger problems so that you have a bigger sense of what is going on and more deep practice.1144

So, let us see. In this particular case, I am going to go ahead and put equation 1 and equation 2 together, so I am going to take equation 1.1155

When I take number 1 - number 2, I get the following... I get x - y = λ1 × x - y, okay?1169

When I take equation 2 and I add it to equation number 3, I end up with y + z = λ1 × y - z.1183

In this case what I have done is I have essentially reduced this so that now all I have is λ1 and I do not have to worry about λ2.1201

Okay. So let us take a look at the first one. So, x - y = λ1 × x - y is satisfied in 2 cases. Excuse me.1209

It is satisfied in 2 cases. The 2 cases are when x does not equal y, λ1 = 1, and the other case is when x = y, and λ1 does not equal 1. Let us take these cases individually and see what we can do.1223

Okay. So, let us do case 1 first when x is not equal to y. So, case 1.1249

So, x does not equal y. When x does not equal y, then this implies that λ = 1, right? You are just taking x - y/x - y, you are getting λ1 = 1.1257

Okay. If we put this in for one... okay, now let us take the second equation. So, put this into y + z = λ1 × y - z.1274

We get y + z equals y - z, which implies that 2z = 0, which implies that z = 0. Now, since z = 0, let us go ahead and use equation number 4. 1303

That gives x2 + y2 = 0, which implies that x = 0 and that y = 0.1322

Okay. So, now we have 0, 0, 0. We found a certain point. However, the point (0,0,0) does not satisfy equation number 5. 1338

Again, all of the equations have to be satisfied, so, we have taken these three equations, we have taken 1, 2, 3, & 4, we have managed to satisfy 1, 2, 3, & 4, but it does not satisfy equation 5.1355

So, (0,0,0) is not a viable point here. So, (0,0,0) is not a viable point to consider, so we have taken care of that case. Now, let us go ahead and deal with case 2.1367

In case 2, we said that x actually equals y, and that λ1 does not equal 1. 1382

If we put equations 4 and 5 together, let me see here, so if x = y what we end up with is the following.1393

We get z2 = x2 + y2, that is the equation 4, and z is equal to 1 + 2x, because it is 1 + x + y, y = x, therefore it is 1 + 2x.1408

Now, let me go ahead and square this, so when I square this and put this into here, I end up with 1 + 2x2 = x2 + y2, but x = y, so it equals x2 + x2, so I am going to get 1 + 4x + 4x2 = 2x2.1427

Then, I am going to move everything over and basically end up with a quadratic here, so I am going to end up with 2x2 + 4x + 1 = 0.1460

Notice, in this case I have used both equation 4 and equation 5 so that is satisfied, and this case of x = y, λ1 not equal to 1, that comes from the other equations so all of the other equations are satisfied here.1475

When I solve this, I cannot factor it so I have to use the quadratic equation. I am going to end up with x = -4 + or - sqrt(8)/4, which equals -1 + or - sqrt(2)/2. That equals x.1486

Well, it also equals y, so, y = the same thing because x = y. Alright.1510

Then, when we go ahead and we put in for z, when we put this, this x value up into this equation to solve for z, we are going to end up with the following z value.1519

You are going to get z = -1 + or - sqrt(2). I will not go ahead and do that arithmetic, but that is how I go ahead and get it. 1532

So, I have x, I have y, and we have z. Now let us go ahead and finish this up, so let us see here.1543

Let us write it this way. x = -1 + sqrt(2)/2, and it also equals -1 - sqrt(2)/2.1552

y is equal to the same thing. -1 + sqrt(2)/2, and -1 - sqrt(2)/2.1569

Our z value is equal to -1 + sqrt(2) - 1 - sqrt(2), so these are our 2 points that we found, okay? These are our 2 points. Let us call this one p1, and p2, these (x,y,z), these (x,y,z), these 2 points satisfy all of the conditions.1578

Now, okay. Now let us go ahead and write out the conclusions for this. Now, we are looking at either an ellipse or a hyperbola.1602

Statistically, when you do this you are going to get an ellipse or a hyperbola. The chances of getting a circle are kind of... you know... that would just mean a plane that is straight across like that, you have a cone and it just cuts across and cuts it in a circle. 1627

That is not what is happening here so you are definitely looking at either an ellipse or a hyperbola, okay?1642

Now, if it is an ellipse, if we have an ellipse, then, p1 is the nearest point -- in other words, it is the minimum of the function f -- is the nearest point, and p2 is the farthest point.1648

The way you get that -- again, once you get your points, you put these points back into x. Well, our function x, our function I am sorry was x2 + y2 + z2. 1677

You put in this squared + this squared + this squared, and then you get a certain value for f at this point, then you put in this squared + this squared + this squared. 1685

When you do that you will find out that this number is less than this number. That is how we do it... is farthest because f(p1) is less than f(p2).1696

So, if this is an ellipse, this is the farthest point, this is the nearest point. Okay.1709

If a hyperbola -- there is a lot to consider here, it is kind of interesting, it would be nice if we could just plug our numbers in and get a number and be done, but we have to actually consider what is happening.1716

If a hyperbola, then there is no furthest point, then there is no farthest point, because a hyperbola is asymptotic.1728

In other words, it goes on into -- you know -- a particular direction infinitely asymptotic.1756

So, p1 and p2 are the two nearest points. This is why earlier on I decided to use the word extrema instead of max/min because you never know what situation you are dealing with.1767

It is not that you are maximizing or minimizing a function, yes, within a given situation you might be maximizing or minimizing, but in this case we do not know whether we are dealing with an ellipse or a hyperbola.1784

If it is an ellipse, then yes, we have a max value and a min value, in other words a farthest point and a nearest point. 1792

If it is a hyperbola, we do not have a maximum value. What we have is 2 points that satisfy all of the conditions, but they happen to be the two points that are nearest to the origin. 1800

There are physical things to consider, there are other things to consider than just max/min, it is not just max/min -- the two nearest points. Okay.1809

Now, so, that is it. This was just an example of a Lagrange multiplier example using 2 constraints, using 2 particular functions that had to be satisfied. 1820

In this case our function that had to be satisfied was our distance function, and it had to satisfy the equation of a cone, and it had to satisfy the equation of the plane. That is it.1830

Everything else was exactly the same. You are solving the series of simultaneous equations. Once you actually get to a point where you get your points, you have to stop and actually think about what it is that is actually going on.1839

This is why we did this final analysis. We have the possibility of an ellipse, we have the possibility of a hyperbola, I am actually going to leave it to you to actually think about which one it actually is.1851

To see if you can actually use the equations that are given, the g1, the g2, and the f to figure out are you getting an ellipse, or are you getting a hyperbola.1862

So, I will leave it with this: see if you can figure out which -- and I will give you a hint, try converting the functions into cylindrical coordinates, if you happen to be familiar with cylindrical coordinates.1872

If not, do not worry about it. It is not a big deal. We do not actually have to figure out, all we wanted to do was find out which was furthest, which was closest, we found our two points, the actual problem itself is satisfied.1892

Okay. Thank you very much for joining us here at educator.com. We will see you next time. Bye-bye.1903