You operate a gaming website, where users must pay a small fee to log on. When you charged $4 the demand was 540 log-ons per month. When you lowered the price to $3.50, the demand increased to 810 log-ons per month.a) Construct a linear a linear demand function for your website and hence obtain the monthly revenue R as a function of the log-on fee x. R ( x ) = ?b) Your internet provider charges you a monthly fee of $10 to maintain your site. Express your monthly profit P as a function of the log-on fee x. P ( x ) = ?Determine the log-on fee you should charge to obtain the largest possible monthly profit. x = $ ?What is the largest possible monthly profit? $ ?

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You operate a gaming website, where users must pay a small fee to log on. When you charged $4 the demand was 540 log-ons per month. When you lowered the price to $3.50, the demand increased to 810 log-ons per month.a) Construct a linear a linear demand function for your website and hence obtain the monthly revenue R as a function of the log-on fee x.  R  (  x  )  =  ?b) Your internet provider charges you a monthly fee of $10 to maintain your site. Express your monthly profit P as a function of the log-on fee x.  P  (  x  )  =  ?Determine the log-on fee you should charge to obtain the largest possible monthly profit.  x  =      $    ?What is the largest possible monthly profit?      $    ?

Marisa Colon · Accepted Answer

Step 1- when you charged $4, demand was 540 log-ons per month,- when you charged $3.50, demand was 810 log-ons per month.a) Consider linear demand function,       D    q    =  m  q  +  b, where q is quantity.Here we have two ordered pairs (540,4) and (810,3.5) first, we calculate slope of demand function.  m  =            3.5      −      4              810      −      540        =            −      0.5        270    =  −  0.00185  ∴      D    q    =  −  0.00185  q  +  bfor (540,4), we have,   4  =  (  −  0.00185  )  540  +  b  4  =  −  0.99  +  b  ⇒  b  =  4.99      D    1    =  −  0.00185  q  +  4.99    i  .  e  .  ,  R  (  x  )  =  −  0.00185  x  +  4.99 is monthlyrevenue R as function of log-on fee x.Step 2b) If internet provides charges a monthly fee of $10, so, here   R  (  x  )  =      $    10, we have to find x.  10  =  −  0.00185  x  +  4.99  ⇒  x  =      $    2708monthly profit P as function of log-on fee x.is   P  (  x  )  =            4.99      −      R      (      x      )        0.00185    =  2697.29  −  540.54  R  (  x  )  ∴  P  (  x  )  =  2697.29  −  540.54  R  (  x  )If   R  (  x  )  =  0, largest possible monthly profit P(x) will be maximum.  largest     P  (  x  )  =      $    2697.29

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