Economics based on investment

Oct 2013
6
0
Hi I am in dia need of help on the following question can someone please help.

A retired junk yard operator, Edsel Studebaker, recently died and left his entire fortune of $12,300,000toatrusttoestablishajunkcarmuseum. Thetrusteesmustdecidehowtospendit. Landcanbepurchasedfor$180,000now. Themuseumbuildingwillrequire30,000squarefeet of administrative and work space. Each car displayed will require an additional 1000 square
feet. Planning will cost $100,000 immediately. Building construction will cost $100 per square foot and will take the next two years (assume equal costs each year.) Cars will cost $500 each (they're junk, remember?) and they will be purchased during the second year of construction. Annual operation ofthe museum will cost $200,000 plus $30 per car. Ifthe funds are invested at 9% per year and the museum is to exist forever, how many cars should the trustees plan to purchase?

Any help would be great, cheers.
 
Oct 2013
6
0
Sorry the sentence above I've left out a series of spaces. Here is the question again for easier reading.


A retired junk yard operator, Edsel Studebaker, recently died and left his entire fortune of $12,300,000 to a trust to establish a junk car museum. The trustees must decide how to spend it. Land can be purchased for $180,000 now. The museum building will require 30,000 square feet of administrative and work space. Each car displayed will require an additional 1000 square
feet. Planning will cost $100,000 immediately. Building construction will cost $100 per square foot and will take the next two years (assume equal costs each year.) Cars will cost $500 each (they're junk, remember?) and they will be purchased during the second year of construction. Annual operation ofthe museum will cost $200,000 plus $30 per car. Ifthe funds are invested at 9% per year and the museum is to exist forever, how many cars should the trustees plan to purchase?
 

Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
Code:
YR        TRANSACTIONS             INTEREST        BALANCE
 0  12,300,000[1]  -280,000[2]                   12,020,000
 
 1  -1,500,000[3]                 1,081,800      11,601,800

 2  -1,500,000[3]  -100,500C[4]   1,044,162      11,145,962 - 100,500C

 3    -200,000[5]       -30C[6]           ?               ?

 Solve for C:
 .09(11145962 - 100500C) = 200000 + 30C
 C = 88.499... so 88 cars

 0  12,300,000     -280,000                      12,020,000
 
 1  -1,500,000                    1,081,800      11,601,800

 2  -1,500,000   -8,844,000       1,044,162       2,301,962

 3    -200,000       -2,640         207,178       2,306,500

 4    -200,000       -2,640         207,585       2,311,445
C = number of Cars.

[1] from estate
[2] right away expenses: 180,000 + 100,000
[3] admin sq. footage: 30,000 @ $100 = 3,000,000: 1,500,000 year1 and year2
[4] per car purchase: 1,000 sq.ft @ $100 = 100,000 + 500 = 100,500
[5] annual operating expenses, beginning year3
[6] per car annual expense, beginning year3

Note that fund increases slightly: due to 88 cars instead of 88.49...
Enough for a huge xmas party each year :p
 
Oct 2013
6
0
Thank you so much Denis this is great!!

Just one question, to solve for c how did you rearrange the equation to do so?

Cheers.
 

Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
Kaiso said:
Just one question, to solve for c how did you rearrange the equation to do so?
Huh? You mean your teacher gives you this involved problem,
but has not teached how to solve such a basic equation? :shock:
Or did you skip classes in order to go to the poolroom? :p

Ok; the equation looks like this:
5(10 - 2c) = 4 + 3c
Multiply what's inside brackets by 5:
50 - 10c = 4 + 3c ; continue...
13c = 46
c = 46/13

OK?
 
Oct 2013
6
0
Hahahaha Algebra always confused me with the chopping and changing either side of the =, :?

Thank you heaps for your feed back, I appreciate it.
 
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