A box with an open top has vertical sides, a square bottom, and a volume of 4 cubic meters. If the box has the least possible surface area, find its dimensions.

Question

A box with an open top has vertical sides, a square bottom, and a volume of 4 cubic meters. If the box has the least possible surface area, find its dimensions.

What is the height?

What is the length of the base?

Answer 1

volume =v = 4 m^3= hs^2

s^2=v/h=4/h

s =2/sqrh

surface area = A = 4(hs)+s^2= 4(h(2/sqrh)+ 4/h = 8sqrh +4/h

A' = 4/sqrh - 4/h^2 =0

4hsqrh -4 =0

h^1.5 = 1

h = 1 meter = height

s= 2 meters = side of the bottom

Answer 2

Let

L=length

W=width

H=height

Given

LWH=4

L=W

Then

Surface Area = A = LW + 2LH + 2WH

W²H = 4

W² = 4/H

A = W² + 2WH + 2WH

   = W² + 4WH

   = 4/H + 4(2/√H)H

   = 4/H + 8√H

A_H’ = -4/H² + 8(1/2)H^-1/2

A_H’ = -4/H² + 4H^-1/2

Get the critical point by setting A_H’ = 0.

-4/H² + 4H^-1/2 = 0

-4 + 4H^3/2 = 0

-1 + H^3/2 = 0

H^3/2 = 1

H = 1

W² = 4

W = 2

L = W = 2

Height = 1

Length = 2

Proof that the critical point of H=1 is a minimum

A_H’’ = 8/H³ – 2H^-3/2

A_H’’(1) = 6 > 0 => minimum