A box has a bottom with one edge 7 times as long as the other. If the box has no top and the volume is fixed at V, what dimensions minimize the surface area?

Answer 1

let the 3 dimensions be
a, b, and c where a and b are the bottom edges and c is the height.
v is the volume
s is the surface area
a = 7b
v = abc
= 7b^2c
or rearranging,
c = v/7b^2
s = ab + 2ac + 2bc
= 7b^2 + 14bc + 2bc
= 7b^2 + 16bc
= 7b^2 + 16bv/7b^2
= 7b^2 + (v16/7)/b
set the 1st derivative equal to zero,
0 = 14b - (v16/7)/b^2
(v16/7) = 15b^3
b=(v16/(7*15))^(1/3)

Answer 2

what are the complete dimensions?

Answer 3

I agree with to FredR, up until
0 = 14b - (16v/7)/b^2

The next step should be
b^3 = 16v/(14*7)= (8/49)v
b = 2*(v/49)^1/3 = 0.5466 v^1/3
Then a = 7b = 3.8264 v^1/3
and c = v/(7b^2) = 0.4781 v^1/3

Answer 4

Oops. Thanks for the correction.