Question

find the dimensions of the rectangular box,open at the top,of maximum capacity whose surface is 432 sq/cm

Answer 1

Dimension of Box = 12 * 12 * 6   cm with maximum capacity whose top is open & surface area = 432 cm²

Step-by-step explanation:

Let say Dimension of the box = L , B & H

Surface Area = LB + 2(L + B)H  = 432

Volume = LBH

Capacity is maximum when volume is maximum.

For Maximum Volume  for given Surface Area

Length = Breadth

L = B  = x

=> x² + 4xh =  432

=> h = (432 - x²)/4x

Volume = x²h

= x² (432 - x²)/4x

= x(432 - x²)/4

= 108x  - x³/4

V =  108x  - x³/4

dV/dx = 108 - 3x²/4

putting dV/dx = 0

=> 3x²/4 = 108

=> x² = 36 * 4

=> x = 12

d²V/dx² = -6x/4  is -ve hence Volume is maximum

Length = 12

Breadth = 12

h = (432 - x²)/4x   =  (432 - 12²)/4(12)

=> h = 6

Dimension of Box = 12 * 12 * 6   cm

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