find the dimensions of the rectangular box open at the top of maximum capacity whose surface area is 108 sq.inches
Answers
Dimension of Box = 6 * 6 * 3 inch of the rectangular box open at the top of maximum capacity whose surface area is 108 sq.inches
Step-by-step explanation:
Let say Dimension of the box = L , B & H
Surface Area = LB + 2(L + B)H = 108
Volume = LBH
Capacity is maximum when volume is maximum.
For Maximum Volume for given Surface Area
Length = Breadth
L = B = x
=> x² + 4xh = 108
=> h = (108 - x²)/4x
Volume = x²h
= x² (108 - x²)/4x
= x(108 - x²)/4
= 27x - x³/4
V = 27x - x³/4
dV/dx = 27 - 3x²/4
putting dV/dx = 0
=> 3x²/4 = 27
=> x² = 36
=> x = 6
d²V/dx² = -6x/4 is -ve hence Volume is maximum
Length = 6
Breadth = 6
h = (108 - x²)/4x = (108 - 6²)/4(6)
=> h = 3
Dimension of Box = 6 * 6 * 3 inches
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