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Allisa [31]
3 years ago
7

A. Following are the transactions of a new company called Pose-for-Pics.

Business
1 answer:
stich3 [128]3 years ago
4 0

Answer:

The August 31 trial balance is a debit and credit amount of $16,986

Explanation:

The journal entries for the following transactions is as follows;

  General Journal                          Debit                              Credit

1. Cash                                            $6,500

  Photography equipment            $33,500

  Common stock                                                                 $40,000

2. Prepaid insurance                       $2,100

   Cash                                                                                   $2,100

3. Office supplies                             $880

   Cash                                                                                    $880

4. Cash                                              $3,331

   Photography fees                                                              $3,331

5. Utilities expense                           $675

   Cash                                                                                     $675

August 31 trial balance                     $16,986                         $16,986  

The August 31 trial balance is a debit and credit amount of $16,986

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Solution:

1)

Profit function of customised bicycle customers, P1 = d1*(p1-c)

= (11000-25p1)*(p1-160)

= 11000p1-1760000-25p1^2+4000p1

= -25p1^2+15000p1-1760000

In order to the profit maximizing price, equate the first order derivative of profit function to 0

dP1/dp1 = d(-25p1^2+15000p1-1760000)/dp1 = 0

=> -50p1+15000 = 0

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Profit function of price sensitive customers, P2 = d2*(p2-c)

= (11000-45p2)*(p2-160)

= 11000p2-1760000-45p2^2+7200p2

= -45p2^2+18200p2-1760000

In order to the profit maximizing price, equate the first order derivative of profit function to 0

dP2/dp2 = d(-45p2^2+18200p2-1760000)/dp2 = 0

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Price to be charged for customised segment = $ 300

Price to be charged for price sensitive segment = $ 202.22

------------------------------------------------------------

2)

Considering single price, p

Total profit from both segments, P = (d1+d2)*(p-160)

= (11000-25p+11000-45p)*(p-160)

= (22000-70p)*(p-160)

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In order to the profit maximizing price, equate the first order derivative of profit function to 0

dP/dp = d(-70p^2+33200p-3520000)/dp = 0

=> -140p+33200 = 0

=> p2 = 237.14

3)

This is solved by Solver as follows:

[ Find the attachments ]

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