In space, the intersection of two planes is
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<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
Substituting the value of H in (1), we get:
Now, to minimize the area put :
Putting the value of S = 15.874 cm in the value of H , we get:
Hence, the dimensions of the given rectangular box are:
L = 15.874 cm
B = 15.874 cm
H = 7.8937 cm