14.8
5.
f (x, y) = y 2 − x2 ; 14 x2 + y 2 = 1
solve ∇f = λ∇g
(−2x, 2y) = λ( x2 , 2y)
⇒ −2x = λx
2
⇒ λ = −4 or x = 0
if λ = −4 then y = 0, x = ±2, (x, y) = (±2, 0) ⇒ f (±2, 0) = −4
if x = 0 then y = ±1, λ = ±1, (x, y) = (0, ±1) ⇒ f (0, ±1) = 1
minimum of f is −4 and maximum of f is 1
12.
f (x, y, z) = x4 + y 4 + z 4 ; x2 + y 2 + z 2 = 1
solve ∇f = λ∇g
(4x3 , 4y 3 , 4z 3 ) = λ(2x, 2y, 2z)
⇒ 4x3 = 2λx ⇒ λ = 2x2 or x = 0
if λ = 2x2 then we have (x, y, z) = (1, 0, 0), (± √12 , 0, ± √12 ), (± √13 , ± √13 , ± √13 ) or (± √12 , ± √12 , 0)
if x = 0 then we have (x, y, z) = (0, 1, 0), (0, ± √12 , ± √12 ) or (0, 0, 1)
f = x4 + y 4 + z 4 takes values 1, 21 , 13 on these points.
So minimum of f is 31 and maximum of f is 1.
16.
f (x, y, z) = 3x − y − 3z; x + y − z = 0, x2 + 2z 2 = 1
solve ∇f = r∇g + s∇h
(3, −1, −3) = r(1, 1, −1) + s(2x, 0, 4z)
x = 2s , z = −1
s
√
∵ x2 + 2z 2 = 1 ⇒ s = ± 6
−2 3 1
, −1
) or ( √
⇒ (x, y, z) = ( √26 , −3
, , )
6
6
6 6 6
12
−12
f takes values √6 and √6
12
√ and maximum of f is √
So minimum of f is −12
.
6
6
18.
f (x, y) = x2 + y 2 + z 2 ; x − y = 1, y 2 − z 2 = 1
solve ∇f = r∇g + s∇h
(2x, 2y, 2z) = (r, −r + 2sy, −2sz)
1
, and z = 0 or s = −1
⇒ x = 2r , y = s−2
if s = −1 we have y = −1
x = 32 ⇒ z 2 = 19 − 1 a contradiction
3
if z = 0 we have (x, y, z) = (2, 1, 0) or (0, −1, 0)
f = x2 + y 2 + z 2 takes values 5 and 1 on these points.
So minimum of f is 1 and maximum of f is 5.
20.
f (x, y) = 2x2 + 3y 2 − 4x − 5, x2 + y 2 ≤ 16
find critical points:
∇f = (4x − 4, 6y), critical point (x, y) = (1, 0), f (1, 0) = −7
find extreme values on boundary:
solve ∇f = λ∇g
(4x − 4, 6y) = λ(2x, 2y)
⇒ λ = 3 or y = 0
1
√
if λ = 3 , (x, y) = (−2, ±2 3)
if y = 0, (x, y) = (±4, 0)
f takes values 47 and 11 on these points.
So minimum of f is −7 and maximum of f is 47.
22.
(a)
solve ∇f = λ∇g
(2, 3) = λ( 2√1 x , √1y )
√
√
⇒ (x, y) = (9, 4) , f (9, 4) = 30, which is the only extreme value of f subject to x + y = 5.
(b)
f (25, 0) = 50, Yes, it’s larger than f (9, 4) = 30
(c)
(d)
The method of Lagrange multipliers identifies points where the level curves of f have common
tangent line with curve g. But this may not occur at an endpoint of g, while an extreme value
can occur at such points.
(e)
√
√
f (9, 4) = 30 is the minimum value of f subject to x + y = 5
23.
(a)
solve ∇f = λ∇g
(1, 0) = λ(4x3 − 3x2 , 2y)
⇒ (x, y) = (1, 0), f (1, 0) = 1
(b)
want to prove f (x, y) ≥ 0, ∀(x, y) ∈ R2 satisfies y 2 + x4 − x3 = 0
if f < 0 at (x0 , y0 ) then x0 < 0
y02 + x40 − x3 = 0 ⇒ y02 = x30 − x40 < 0 a contradiction
thus f (0, 0) = 0 is the minimum value.
And ∇f (0, 0) = (1, 0) 6= λ(0, 0) = λ∇g(0, 0) for all λ ∈ R
(c)
∇g(0, 0) = (0, 0) in this case, but the method of Lagrange multipliers require the property that
∇g 6= 0 on the surface g = k.
2
43.
Let f (x, y, z) = x2 + y 2 + z 2 , want to find extreme values subject to g = x + y + 2z = 2 and
h = x2 + y 2 − z = 0.
solve ∇f = r∇g + s∇h
(2x, 2y, 2z) = (r + 2sx, r + 2sy, 2r − s)
2x = r + 2sx, 2y = r + 2sy, ⇒ 2(x − y) = 2s(x − y)
⇒ s = 1 or x = y
, a contradiction.
if s = 1, then x2 + y 2 = −1
2
so we must have x = y.
⇒ z = 1 − x, and 2x2 = z
⇒ x = 1 or −1
2
(x, y, z) = ( 21 , 21 , 12 ) or (−1, −1, 2)
f ( 12 , 21 , 12 ) = 43 and f (−1, −1, 2) = 6
Thus the nearest point is ( 12 , 12 , 21 ), the farthest point is (−1, −1, 2)
47.
(a)
solve ∇f = λ∇g
(x2 x3 ...xn , x1 x3 ...xn , ...x1 ...xn−1 ) = λ(1, 1, ..., 1)
Assume none of xi , i = 1, ..., n is zero(otherwise f (x1 , ..., xn ) = 0).
⇒ x1 = x2 = ... = xn = nc
f ( nc , ..., nc ) = nc which is the maximum value of f .
(b)
x1 + x2 + ... + xn
= nc in this case.
n
√
And from part (a) we have n x1 x2 ...xn ≤ nc .
thus we have
√
x1 + x2 + ... + x + n
n
x1 x2 ...xn ≤
n
3