Question

Q-3 x, y and z are positive real numbers such that xyz = 32. Find the minimum value of x^2 + 4xy + 4y^2 + 2z^

Answer 1

Answer:

I am hoping that last term is z^2

then

keep x = 2

y = 1

z = 16

Ans will be

4+8+4+16^2 = 272

Answer 2

Step-by-step explanation:

here you go thank you