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The sum of the surface areas of a rectangular parallelepiped with sides x, 2x, and and a sphere is given to be constant. Prove that the sum of their volumes is minimum if x is equal to three times the radius of the sphere. Also, find the minimum value of the sum of their volumes.

0 3 Views | Posted 3 months ago
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    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Long Answer Type Question as classified in NCERT Exemplar

    Sol:

       

    L e t     ' r '     b e     t h e     r a d i u s     o f     t h e     s p h e r e . ∴     S u r f a c e     a r e a     o f     t h e     s p h e r e = 4 π r 2 V o l u m e     o f     t h e     s p h e r e = 4 3 π r 3 T h e     s i d e s     o f     t h e     p a r a l l e l o p i p e d     a r e     x ,   2 x     a n d     x 3 ∴         I t s     s u r f a c e     a r e a = 2 [ x × 2 x + 2 x × x 3 + x × x 3 ]                                                                                       = 2 [ 2 x 2 + 2 x 2 3 + x 2 3 ] = 2 [ 2 x 2 + x 2 ]                                                                                         = 2 [ 3 x 2 ] = 6 x 2 V o l u m e     o f     t h e     p a r a l l e l o p i p e d = x × 2 x × x 3 = 2 3 x 3 A s     p e r     t h e     c o n d i t i o n s     o f     t h e     q u e s t i o n , surface  area  of  the  parallelopiped+surface  area  of  the  sphere=constant ⇒                       6 x 2 + 4 π r 2 = K ( constant )           ⇒ 4 π r 2 = K − 6 x 2 ∴                                                                 r 2 = K − 6 x 2 4 π                                                                                                                                           … ( i ) N o w     l e t                                     V = V o l u m e     o f     t h e     p a r a l l e l o p i p e d + V o l u m e     o f     t h e     s p h e r e ⇒                                                               V = 2 3 x 3 + 4 3 π r 3 ⇒                                                               V = 2 3 x 3 + 4 3 π [ K − 6 x 2 4 π ] 3 / 2                   [ F r o m     e q . ( i ) ] ⇒ &thi

    S q u a r i n g     b o t h     s i d e s ,     w e     g e t ⇒               4 π x 2 = 9 ( K − 6 x 2 )                 ⇒ 4 π x 2 = 9 K − 5 4 x 2 ⇒                   4 π x 2 + 5 4 x 2 = 9 K ⇒                 K = 4 π x 2 + 5 4 x 2 9                                                                       ( i i ) ⇒                 2 x 2 ( 2 π + 2 7 ) = 9 K ∴                                                           x 2 = 9 K 2 ( 2 π + 2 7 ) = 3 N o w     f r o m     e q . ( i )     w e     h a v e                                                               r 2 = K − 6 x 2 4 π ⇒                                                     r 2 = 4 π x 2 + 5 4 x 2 9 − 6 x 2 4 π ⇒                                                     r 2 = 4 π x 2 + 5 4 x 2 − 5 4 x 2 9 × 4 π = 4 π x 2 9 × 4 π = x 2 9         ⇒ r = x 3         ∴ x = 3 r N o w     w e     h a v e               d V d x = 2 x 2 − 3 x ( K − 6 x 2 ) 1 / 2

                                              = 1 2 − 3 4 K π + 5 4 K − 1 0 8 K 4 π + 5 4                                           = 1 2 − 3 [ 4 K π − 5 4 K 4 π + 5 4 ]                                           = 1 2 − 3 [ 4 K π − 5 4 K . ]                                           = 1 2 − 6 [ 2 π − 2 7 . ]                                           = 1 2 + 6 K [ 2 7 − 2 π . ] > 0                           [ ? 2 7 − 2 π > 0 ] ∴  d2Vdx2>0  So,  it  is  minima. Hence,  the  sum  of  volume  is  min

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If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to..........
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Raj Pandey

y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406|

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If the tangent to the curve, y = f(x) = xlog? x, (x > 0) at a point (c, f(c)) is parallel to the line - segment joining the points (1,0) and (e, e) then c is equal to:
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Raj Pandey

f' (c) = 1 + lnc = e/ (e-1)
lnc = e/ (e-1) - 1 = (e - (e-1)/ (e-1) = 1/ (e-1)
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The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 – 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to:
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C D = ( 1 0 + x 2 ) 2 − ( 1 0 − x 2 ) 2 = 2 1 0 | x |

Area 

= 1 2 × C D × A B = 1 2 × 2 1 0 | x | ( 2 0 − 2 x 2 )

⇒ 1 0 − x 2 = 2 x

3x2 = 10

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3k2 = 10

Let F1(A,B,C) = ( A ∧ ∼ B ) ∨ [ ∼ C ∧ ( A ∨ B ) ] ∨ ∼ A     a n d     F 2 ( A , B ) = ( A ∨ B ) ∨ ( B → ∼ A )  be two logical expressions. Then:
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Vishal Baghel

By truth table

So F1 (A, B, C) is not a tautology

Now again by truth table

So      F2 (A, B) be a tautology.

The triangle of maximum area that can be inscribed in a given circle of radius ‘r’ is:
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Vishal Baghel

From option let it be isosceles where AB = AC then

x = r 2 − ( h − r ) 2            

=  r 2 − h 2 − r 2 + 2 r h

x = 2 h r − h 2 . . . . . . . . ( i )

 Now ar ( Δ A B C ) = Δ = 1 2 B C × A L

Δ = 1 2 × 2 2 h r − a 2 × h        

then  x = 2 × 3 r 2 × r − 9 r 2 4 = 3 2 r         f r o m ( i )

⇒ B C = 3 r    

So A B = h 2 + x 2 = 9 r 2 4 + 3 r 2 4 = 3 r .

Hence Δ be equilateral having each side of length 3 r .

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