Question

Please explain the concepts in your answer. Thank you.

If x, y, z are three real numbers such that x4 + y4 + 24 = 243, what is the largest possible value of xyz? Justify your answer.

Answer 1

Given Condition, x + y2 + 2 = 243 Let, 9 = x+y+z9–243.00 = 0 Let f(x, y2) = xyz For Lagrange multiplic method F = f t gd F = xyz + gde F = xyz + d(x + y + z = 243) If = 1) yz + 4x² = y2 + 4x² DE 23 + 4yd at = xy + 420 4 4 4

dd . dd of 0 + ) (2 +++ 29 –243) of = x++y + 29 – 243 For largest value of f = xyz See = , D = =, f Y&+488 = o, X+4% =0 xy +42d=0 48d = -92, 95%d=-2x, 42 N = -xy and = 72 = 2 3 4 x4 - 49 - [3=9 g [or=-9] Dividing - 22

Similiarly, 4931 – – XZ 4232 - xy = 773 = 5 Therefore i 19=2], or y=-) x=y=2 , 10c=y=-2 / axd=-92 xd = -xx lor 4x² d = -x-(-x) 400 = -x of 10 = x² xc² (4x + 1) = 0 XXd + 1 = 0 of deb x² (4 x + d) = 0 x = -1 (= x=y= 2 g =y=-2, x= - = -2 y = -24; 2=-4 2-4 40

Now For df =0 and X=Y=2= - 4 x²+49 + 24 – 243 = 0 Cad) + ( 2 ) + ( 4 ) = 243 °C tot + a) = 245 to = 94x $1 to = (4x2) I - 4x3 d = te

12 x=y=2=-2 / d x = -3 g=y= 2 = -3 f = xyz = 63) (-3) (-3) =-27 f = xyz =-27 For x=y=2= .: 2 = 1 / For at = 0 and x=y = 1/2 x² + y + z² – 243 20 (A) + ( 1 ) + ( 1 ) - 243 = 0 94 x 2 = 243 d = 1/2

For date x= - = ext=3 y = x = 3 2--- x+ = -3 3 – xy2 = $(3) (3) (-3) = -24 Fol d=-12 x=y=ofd = ax = -3 zada = 3 as :f =X42 = (-3) (-3) (3) = 27 - T - 9 -

7 = xyz & has largest value for egn (3 amoney equation , ③ and ② Barat = xyz = 27 : Largest value ab slyz = 27.