Show that if xyz=8, then the minimum value of the product (x+1)(y+1(z+1) is 27?

Question

How do prove this thing using partial derivatives or by Method of Lagrange Multiplier for Maxima & Minima Under Constraints?

Answer 1

Simplify by rewriting it like this

(x+1)(y+1)(z+1) = 1+x+y+z+xy+yz+zx+xyz
=9+x+y+z+xy+yz+zx

Then using a lagrange multiplier c we want to minimise

9+x+y+z+xy+yz+zx+c*(8-xyz)

differentiating wrt x,y and z gives the set of three equations

1+y+z=cyz
1+x+z=cxz
1+x+y=cxy

multiply the first eqn by x, the second by y and the third by z we can replace xyz with 8 in each and see that

8c=x(1+y+z)=y(1+x+z)=z(1+x+y)

taking these in pairs we get three eqns like

x(1+y+z)=y(1+x+z)

or

x-y=z(y-x)

so the solution needs to satisfy

x=y or z=-1
and
y=z or x=-1
and
z=x or y=-1

The only consistent solns are when either two of x,y,z are -1 and the other is 8, or when x=y=z=2. The first cases are minima, whereas the last is actually a maximum.

Answer 2

If x=2, y=2, and z=2 then (2+1)(2+1)(2+1)=27

Answer 3

x=2,y=2,z=2
xyz=8
(2+1)(2+1)(2+1)=27

Answer 4

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Answer 5

2x2x2=8, Therefore, x,y, and z are "2"

Hence, (2+1)(2+1)(2+1)=27

(3*3*3)=27

Answer 6

its wrong. you are mistaken
(-1)(-1)(8)=8
(-1+1)(-1+1)(8+1)=0
hahaha