JEE Mathematics/ Lakshmikanta Satapathy/ Application of derivative QA part 12/ JEE question on Maximum volume of a rectangular tank solved with the related cpncepts

1. Physics Helpline
L K Satapathy
Maxima and Minima 6

2. Physics Helpline
L K Satapathy
Application of Derivative 12
Answer :
Question : A rectangular sheet of tin of size 45cm by 24cm is to be made into a box
without top by cutting off square pieces from each corner and then folding up the
flaps. What should be the side of each square such that the volume of the box is
maximum ?
(a) 4cm (b) 5cm (c) 6cm (d) 7cm
45
24
Original length = 45 cm , breadth = 24 cm
Let side of each square cut from corners = x cm
Length of box = (45 – 2x ) cm
24-2x
45 - 2x
xx
Breadth of box = (24 – 2x ) cm
Height of box = x cm
 Volume of box (45 2 )(24 2 )V x x x  

3. Physics Helpline
L K Satapathy
Application of Derivative 12
Correct option = (b)
3 2
( ) (45 2 )(24 2 ) 4 138 1080Let f x x x x x x x     
2 2
( ) 12 276 1080 12( 23 90) 12( 5)( 18)f x x x x x x x         
& ( ) 12(2 23)f x x  
& (5) 12(10 23) 0f    
( ) 0 5 0 18 0f x x or x      
 Volume of box is maximum when side of square = 5 cm
(i) x – 18 = 0  x = 18 [not possible since breadth (24 – 2x ) becomes -ve ]
( ) 5 0 5ii x x   
 f(x) is maximum when x = 5 cm

4. Physics Helpline
L K Satapathy
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