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Sol: (313)
The volume of water in a tank t minutes after it starts draining is
 
2
( ) 350 20V t t L 
First we need to find the derivative of the formula above to find the rate of
change, i.e. the rate at which the tank is draining at any particular minute, t
 
   
 
2
'( ) 350 20
'( ) 350 2 1 20
'( ) 700 20
d
V t t L
dt
V t t L
V t t L
 
  
 

(a) Water draining out after 5 minutes,
 
 
'(5) 700 20 5 min
'(5) 700 15 min
'(5) 10500 min
V L
V L
V L
 



Therefore the water is draining at a rate of10500 minL .
Water draining out after 15 minutes,
 
 
'(5) 700 20 5 min
'(5) 700 15 min
'(5) 3500 min
V L
V L
V L
 



Therefore the water is draining at a rate of3500 minL .
(b) When 5mint  ,
 
 
 
2
2
(5) 350 20 5
(5) 350 15
(5) 350 225
(5) 78,750
V L
V L
V L
V L
 




When 15mint  ,
 
 
 
2
2
(15) 350 20 15
(5) 350 5
(5) 350 25
(5) 8750
V L
V L
V L
V L
 




So the average rate at which water is draining out during the time interval from
5min to 15min is,

 
 
8750 78,750
14,000 min
15 5 min
L
L

  


This means that during the time interval from 5min to15min , the water decreased by
14,000 minL .
Sol: (363)
Profit per day ( )P x is given by,
2( ) 8 0.005 1000P x x x  
(a) Marginal profit is simply derivative of profit function. So
 
 
     
2
2
'( ) 8 0.005 1000
'( ) 8 0.005 1000 1
'( ) 8 1 0.005 2 1000 0
'( ) 8 0.01
d
P x x x
dx
d d d
P x x x
dx dx dx
P x x
P x x
  
  
  
 

So '( )P x is the...