Explanation:Since the thickness is 0.5cm, we need to subtract it twice (for both sides) from each dimension of the box
i.e. L = 18 - 1 = 17
B = 10 - 1 = 9
H = 12 - 1 = 11
Now the cylinder must stand upright and must have maximum volume.
NOTE: Only the smaller dimension of the face can be taken as the diameterCase I: Base of cylinder at face LB
Diameter = B = 9
Height = H = 11
Volume = \(\frac{891π}{4}\)
Case II: Base of cylinder at face BH
Diameter = B = 9
Height = L = 17
Volume = \(\frac{1377π}{4}\)Case I: Base of cylinder at face LH
Diameter = H = 11
Height = B = 9
Volume = \(\frac{1089π}{4}\)
Maximum volume is given for Case II with diameter as 9 and radius = \(\frac{9}{2} = 4.5\)
Col. A: 4.5
Col. B: 4
Hence, option A
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I hope this helps!
Regards:
Karun Mendiratta
Founder and Quant Trainer
Prepster Education, Delhi, Indiahttps://www.instagram.com/prepster_education/