−2x = λx2 ⇒ λ = −4 or x = 0 if λ = −4 then y = 0, x = ±2, (x, y

14.8 5.

f (x, y) = y²− x²; ¹₄x²+ y² = 1 solve ∇f = λ∇g

(−2x, 2y) = λ(^x₂, 2y)

⇒ −2x = ^λx₂

⇒ 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+ y⁴+ z⁴; x²+ y²+ z² = 1 solve ∇f = λ∇g

(4x³, 4y³, 4z³) = λ(2x, 2y, 2z)

⇒ 4x³ = 2λx ⇒ λ = 2x² or x = 0

if λ = 2x² then we have (x, y, z) = (1, 0, 0), (±^√1

2, 0, ±^√1

2), (±^√1

3, ±^√1

3, ±^√1

3) or (±^√1

2, ±^√1

2, 0) if x = 0 then we have (x, y, z) = (0, 1, 0), (0, ±^√1₂, ±^√1₂) or (0, 0, 1)

f = x⁴+ y⁴+ z⁴ takes values 1,¹₂,¹₃ on these points.

So minimum of f is ¹₃ and maximum of f is 1.

16.

f (x, y, z) = 3x − y − 3z; x + y − z = 0, x²+ 2z² = 1 solve ∇f = r∇g + s∇h

(3, −1, −3) = r(1, 1, −1) + s(2x, 0, 4z) x = ²_s, z = ⁻¹_s

∵ x²+ 2z² = 1 ⇒ s = ±√ 6

⇒ (x, y, z) = (^√2

6,⁻³₆ ,⁻¹₆ ) or (^−2√

6,³₆,¹₆) f takes values ^√12₆ and ^−12√₆

So minimum of f is ^−12√

6 and maximum of f is ^√12

6. 18.

f (x, y) = x²+ y²+ z²; x − y = 1, y²− z² = 1 solve ∇f = r∇g + s∇h

(2x, 2y, 2z) = (r, −r + 2sy, −2sz)

⇒ x = ^r₂, y = _s−2¹ , and z = 0 or s = −1

if s = −1 we have y = ⁻¹₃ x = ²₃ ⇒ z² = ¹₉ − 1 a contradiction if z = 0 we have (x, y, z) = (2, 1, 0) or (0, −1, 0)

f = x²+ y²+ z² 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+ 3y²− 4x − 5, x²+ y² ≤ 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,^√1_y)

⇒ (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 y²+ x⁴− x³ = 0 if f < 0 at (x₀, y₀) then x₀ < 0

y²₀+ x⁴₀− x³ = 0 ⇒ y²₀ = 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+ y²+ z², want to find extreme values subject to g = x + y + 2z = 2 and h = x²+ y²− 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

if s = 1, then x²+ y² = ⁻¹₂ , a contradiction.

so we must have x = y.

⇒ z = 1 − x, and 2x² = z

⇒ x = 1 or ⁻¹₂

(x, y, z) = (¹₂,¹₂,¹₂) or (−1, −1, 2) f (¹₂,¹₂,¹₂) = ³₄ and f (−1, −1, 2) = 6

Thus the nearest point is (¹₂,¹₂,¹₂), the farthest point is (−1, −1, 2) 47.

(a)

solve ∇f = λ∇g

(x2x3...xn, x1x3...xn, ...x1...xn−1) = λ(1, 1, ..., 1)

Assume none of x_i, i = 1, ..., n is zero(otherwise f (x₁, ..., x_n) = 0).

⇒ x₁ = x₂ = ... = x_n= _n^c

f (_n^c, ...,_n^c) = _n^c which is the maximum value of f . (b)

x1+ x2+ ... + xn

n = _n^c in this case.

And from part (a) we have √ⁿ

x₁x₂...x_n ≤ _n^c. thus we have

√n

x1x2...xn ≤ x₁+ x₂+ ... + x + n n

3