Question

If xyzw = 81, What is the minimum value of x + y +z +w ?​

Answer 1

Answer:

Step-by-step explanation:

Arthimic mean ≥ geometric mean

A.M ≥ G.M

$\frac{x+y+z+w}{4}\geq \sqrt[4]{xyzw} \\x+y+z+w\geq 4\sqrt[4]{81} \\x+y+z+w\geq 4*3\\x+y+z+w\geq 12$

so the minimum value of x+y+z+w=12

Answer 2

Answer:

x×y×z×w=81

X+Y+Z+W=81

it means = X=2 ,y =3 z=1 and w= 2