Question

A box 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.)

Answer 1

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


 

Answer 2

Dimension of the box is $3\times6\times6$ .

Step-by-step explanation:

Given :

Volume = 108 cubic meters

Solution :

Let 'a' be the height of the box and 'b' be the one side of the base of the box.

Therefore,

Volume = $\rm ab^2$ = 108

$\rm a = \dfrac{108}{b^2}$   ------ (1)

Now surface area of the box is,

$\rm SA = b^2 + 4ab$  ------ (2)

From equation (1) and (2) we get,

$\rm SA = b^2 + 4\times(\dfrac{108}{b^2})\times b$

$\rm SA = b^2 + \dfrac{432}{b}$  ------ (3)

Now for least possible surface area differentiate equation (3) we get,

$\rm \dfrac{d(SA)}{db}=2b -\dfrac{432}{b^2}$   ------ (4)

Now put the equation (4) equal to zero we get,

$\rm 2b - \dfrac{432}{b^2} = 0$

$\rm b^3 = 216$

b = 6 m

Now put the value of b in equation (1) we get,

a = 3 m

Therefore dimension of the box is $3\times6\times6$ .

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