Question

Find the dimensions of an open rectangular box with a square base that holds 2000 cubic cm and is constructed with the least building material possible.

Answer 1

The dimensions of the given rectangular box are:

L  =   15.874 cm  , B  =  15.874 cm   , H = 7.8937 cm

Step-by-step explanation:

Let us assume that the dimension of the square base = S x S

Let us assume the height of the rectangular base = H

So, the total area of the open rectangular box  

= Area of the base +  4 x ( Area of the adjacent faces)

=  S x S  +  4 ( S x H)   = S² +  4 SH   ..... (1)

Also, Area of the box  = S x S x H  =  S²H

⇒ S²H = 2000

$\implies H = \frac{2000}{S^2}$

Substituting the value of H in (1), we get:

$A = S^2 + 4 SH = S^2 + 4 S(\frac{2000}{S^2}) = S^2 + (\frac{8000}{S})\\\implies A = S^2 + (\frac{8000}{S})$

Now, to minimize the area put :

$(\frac{dA}{dS} ) = 0 \implies 2S - \frac{8000}{S^2} = 0\\\implies S^3 = 4000\\\implies S = 15.874 \approx 16 cm$

Putting the value of S  = 15.874 cm in the value of H , we get:

$\implies H = \frac{2000}{S^2} = \frac{2000}{(15.874)^2} = 7.8937 cm$

Hence, the dimensions of the given rectangular box are:

L  =   15.874 cm

B  = 15.874 cm

H = 7.8937 cm