SOLUTION: Prove or disprove: The rectangular solid of minimum surface area with fixed volume is a cube.

Question 389682: Prove or disprove: The rectangular solid of minimum surface area with fixed volume is a cube.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Without loss of generality, let the fixed volume be 1, and the dimensions of the rectangular solid be x, y, z. Then,

= 1, and we want to show whether

--> is true (if it is true, then the solid must have a surface area of at least 6, and the solid of least surface area is a cube).

Applying the AM-GM inequality,

Since , replace it into the right-hand side and simplify:

as desired. The equality of AM-GM occurs when x = y = z, i.e. the rectangular solid is a cube.

RELATED QUESTIONS
Show that the rectangular solid of maximum surface area inscribed in a sphere is a... (answered by richard1234)

prove that the surface area of a cube of side length centimeters is greater that surface... (answered by Alan3354)

find the total surface area of a solid cube of volume... (answered by josmiceli,Gogonati)

A rectangular solid cube with each side measuring is 10cm. What is the volume of the cube (answered by Alan3354)

I need your help with this,please: The surface area of a rectangular solid is... (answered by MathLover1)

I need help with this problem please: The volume of a rectangular solid is... (answered by josmiceli)

A rectangular solid (with a square base) has a surface area of 121.5 square centimeters.... (answered by Aaleia)

A cube is a solid figure with six faces. Each face is a square. The surface area of a... (answered by richwmiller)

Determine the dimensions of a rectangular solid (with a square base) with maximum volume... (answered by ankor@dixie-net.com)