Abox with an open top has vertical sides, a square bottom, and a volume of 108 cubic meters. if the box has the least possible surface area, find its dimensions. (in your answer leave a space between the number and the unit.)

Let x be the height  of the  box and y be the length of one side of the base then:-

V = xy^2 = 108

x = 108/y^2

Surface area = y^2 + 4xy 
S = y^2 + 4y* 108/y^2
S = y^2 + 432/y
Finding the derivative:-
dS/dy  = 2y - 432/y^2  = 0
2y^3 = 432
y^3 = 216
y = 6 

Check if this gives a minimum value:-
second derivative = 2 + 864/y^3  which is positive so  minimum.

V = xy^2 = 108
36y = 108
y = 3

Answer :-  dimensions of the box is 3*6*6 metres

 

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