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CBSE Class 12-science Answered

Some of surface area?
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Asked by alanpeter9611 | 17 Feb, 2019, 02:48: AM
answered-by-expert Expert Answer
If r is radius of sphere, with the given dimensions of cuboid, total surface area A  is given by,
 
A = 2 [ 2x2 + 2x2/3 + x2/3 ] + 4πr2 = C' ...........................(1)
where C' is a constant
 
Eqn.(1) is divided by 4 and simplified as,   (3/2)x2 + πr2 = C  ...............(2)
where C is constant.
 
Volume of cuboid+sphere,   V  = (2/3)x3 + (4/3)πr3  .........................(3)
 
Let us substitute r as a function of x in eqn.(3), using eqn.(2)
 
from eqn.(2), we have,  begin mathsize 12px style r cubed space equals space fraction numerator open parentheses C space minus space begin display style 3 over 2 end style x squared close parentheses to the power of begin display style bevelled 3 over 2 end style end exponent over denominator straight pi square root of straight pi end fraction end style ................................(4)
using the substitution for r3 in eqn.(3), we write, begin mathsize 12px style V space equals space 2 over 3 x cubed space plus space fraction numerator 4 over denominator 3 square root of straight pi end fraction open parentheses C space minus space 3 over 2 x squared close parentheses to the power of bevelled 3 over 2 end exponent space end style   .............................(5)
To minimize the volume, we differentiate volume function V with respect to x and equate it to zero
begin mathsize 12px style fraction numerator d V over denominator d x end fraction space equals space 2 over 3 cross times 3 x squared space plus space fraction numerator 4 over denominator 3 square root of straight pi end fraction cross times 3 over 2 open parentheses C space minus 3 over 2 x squared close parentheses to the power of bevelled 1 half end exponent cross times open parentheses negative 3 over 2 close parentheses 2 x space space equals space 0 end style ...................................(6)
From eqn.(6), we get,  begin mathsize 12px style x squared space equals space fraction numerator 18 C over denominator 27 space plus space 2 straight pi end fraction end style  .....................................(7)
substituting for x2 in eqn.(2) and solving for r2 , we get  begin mathsize 12px style r squared space equals space fraction numerator 2 C over denominator 27 plus 2 straight pi end fraction end style  ................................(8)
from eqn.(7) and eqn.(8), we have x = 3r
 
Minimum volume is calculated by using r = x/3 in eqn.(3).  Vmin = (2/3)x3 + (4/3)π(x3/27) = [ (54+4π)/81 ]x3
Answered by Thiyagarajan K | 17 Feb, 2019, 09:53: PM

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