The storage space required is given by the function P(N) = 4000 √N, where N is the number of boxes used. Find the percentage change in storage if the number of boxes is increased by 1%.
(a) 0.75%
(b) 0.25%
(c) 0.5%
(d) 1%
(e) 2%

• Since 1% increase in N means 1.01N. So new function becomes P'(N)=4000/(1.01)^1/2. Now, percentage change can be obtained by P(N)-P'(N)/ P(N), thus giving us the solution 0.5%. Also sorry for the previous incomplete solution.
• 7 years agoHelpfull: Yes(1) No(2)
• New no. of boxes=101n/100
  increase=4000/n-400000/101n=4000/n*1/101
  ans=4000/n*1/101/4000/n
  =1/101*100
  =1%
• 6 years agoHelpfull: Yes(1) No(0)
• Since 1% increase in N i.e. no.of boxes is equal to 1.01N.So, new function P(N)=4000/(1.01)
• 7 years agoHelpfull: Yes(0) No(0)
• in given problem, CHANGE IN STORAGE ,so u make differential dp(n)/dn =4000 sqrt(N)
  
  AFTER DIFFERENTIAL, 1/2 N^-1/2 where change in storage =dp(n)/dn=1%=0.01
  
  dp(n)/dn=1/2N^-1/2
  
  0.01=1/2N^-1/2
  
  hence after solving N=100
  
  N=1 %
• 7 years agoHelpfull: Yes(0) No(1)