Math 280, Intermediate Calculus
14-8 Lagrange Multipliers
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42. The plane 4x − 3y + 8z = 5 intersects the cone z 2 = x2 + y 2 in an ellipse.
a. Graph the cone, the plane, and the ellipse.
b. Use Lagrange multipliers to find the highest and lowest points on the ellipse.
Here, the two constraints are g (x, y, z) = x + y + 2z − 2 and h (x, y, z) = x2 + y 2 − z. Any critical
point that we find during the Lagrange multiplier process will satisfy both of these constraints,
so we actually don’t need to find an explicit equation for the ellipse that is their intersection.
Suppose that (x, y, z) is any point that satisfies both of the constraints
(and hence is on the ellipse.)
q
2
2
2
Then the distance from (x, y, z) to the origin is given by (x − 0) + (y − 0) + (z − 0) . This
expression (and its partial derivatives) would be cumbersome to work with, so we will find the
extrema of the square of the distance. Thus, our objective function is
f (x, y, z) = x2 + y 2 + z 2
and
∇f
λ∇g
= h2x, 2y, 2zi
= hλ, λ, 2λi
µ∇h = h2µx, 2µy, −µi
Thus the system we need to solve for (x, y, z) is
2x =
λ + 2µx
(1)
2y
=
λ + 2µy
(2)
2z
=
2λ − µ
(3)
x + y + 2z
=
2
(4)
x2 + y 2 − z
=
0
(5)
Subtracting (2) from (1) and factoring gives
2 (x − y) = 2µ (x − y)
so µ = 1 whenever x 6= y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and
λ = 0 into (3) gives us 2z = −1 and thus z = − 21 . Subtituting z = − 21 into (4) and (5) gives us
x+y−3
1
x2 + y 2 +
2
however, x2 + y 2 +
1
2
=
0
=
0
= 0 has no solution. Thus we must have x = y.
Math 280, Intermediate Calculus, 14-8 Lagrange Multipliers
2
Since we now know x = y, (4) and (5) become
2x + 2z
2
2x − z
=
2
=
0
so
z
=
1−x
z
=
2x2
Combining these together gives us 2x2 = 1 − x, so 2x2 + x − 1 = 0 which has solutions x = 21 and
x = −1.
Further substitution yeilds the critical points 12 , 12 , 12 and (−1, −1, 2). Substituting these into
our objective function gives us
3
1 1 1
, ,
=
f
2 2 2
4
f (−1, −1, 2) = 6
√
Thus minimum distance of
(−1, −1, 2).
3
2
occurs at
1 1 1
2, 2, 2
and the maximum distance of
√
6 occurs at