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Naddika [18.5K]
3 years ago
15

PLEASE HELP IMMEDIATELY!!! WILL MARK AS BRAINLIEST

Mathematics
1 answer:
Levart [38]3 years ago
5 0

Answer:

2

Step-by-step explanation:

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Find the dimensions of an open rectangular box with a square base that holds 2000 cubic cm and is constructed with the least bui
Vesnalui [34]
<h3>The dimensions of the given rectangular box are:</h3><h3>L  =   15.874 cm  , B  =  15.874 cm   , H = 7.8937 cm</h3>

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

3 0
3 years ago
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