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Dmitry_Shevchenko [17]
3 years ago
7

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

treme values of the function subject to the given constraint. f (x, y, z) = x4 + y4 + z4; x2 + y2 + z2 = 13
Please find the maximum and minimum value. The answer is NOT 169/3 or 507/9 for the maximum.

Would the answer be: max 169 min 37.5? If so please provide work and explanation still.
Mathematics
1 answer:
olga nikolaevna [1]3 years ago
3 0

Answer:

maximum value , f max = 169 (taking the most out of the 4th power)

minimum value , f min = 169/3 (taking the least out of the 4th power)

Step-by-step explanation:

A) Since both function and restriction are symmetrical with respect to x,y and z, there is no reason for one to be more important than the others and therefore one solution would be x=y=z=λ and thus

x2 + y2 + z2 = 13  → 3λ² = 13 →λ² = 13/3

and f would be

f (x, y, z) = x4 + y4 + z4 = 3λ⁴ = 3*(13/3)²=13²/3=169/3

since x⁴ increases faster than 3*x² , f(x,y,z) would be a minimum

and the maximum value would be obtained taking the most out of x⁴, thus doing 2 coordinates =0 ( can be x=0 and y=0) and

z²= 13

f (x, y, z) = x4 + y4 + z4 = 13² = 169

B) strictly, using Lagrange multipliers

f (x, y, z) = x4 + y4 + z4

g (x, y, z) = x2 + y2 + z2 - 13

F(x,y,z) = f (x, y, z) -λ*g (x, y, z)

such that

Fx (x,y,z)=  fx(x, y, z) -λ*gx (x, y, z) = 0 → 4*x³ - λ*2*x = 0 → 2*x*(2*x² -λ) = 0

thus x=0 or x²= λ/2

Fy (x,y,z)=  fy(x, y, z) -λ*gy (x, y, z)= 0 → 4*y³ - λ*2*y = 0 → 2*y*(2*y² - λ) = 0

thus y=0 or y²= λ/2

Fz (x,y,z)=  fz(x, y, z) -λ*gz (x, y, z)= 0 → 4*z³ - λ*2*z = 0→ 2*z*(2*z² - λ) = 0

thus z=0 or z²= λ/2

g (x, y, z) = 0  → x2 + y2 + z2 = 13 → 3*(λ/2) = 13 → λ=13*2/3

thus  x²=y²=z²= λ/2 =13/3

f min = f (x, y, z) = x4 + y4 + z4 = 3*(13/3)²=169/3

for the x=0 , y=0 → z²= 13

f max = f (x, y, z) = x4 + y4 + z4 = 13² = 169

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