Example 4.3 .1 . ([1]) For the production company "HASHKOM", in the first half of the year, the following data can be extracted : the realized production in the first and the fourth month reduced for the realized production in the third month is 9.000 kg of a product, and the realized production in the first and fifth month is 20.000 kilograms . If the production in the first half of the year took place on the principle of arithmetic progression that is growing, then to calculate the realized production in the first and in the sixth month?
In[]:=
Table[a[1]+(-1+n)d,{n,{1}}]+Table[a[1]+(-1+n)d,{n,{4}}]-Table[a[1]+(-1+n)d,{n,{3}}]==9000
Out[]=
{d+a[1]}9000
In[]:=
Table[a[1]+(-1+n)d,{n,{1}}]+Table[a[1]+(-1+n)d,{n,{5}}]20000
Out[]=
{4d+2a[1]}20000
In[]:=
Solve[{d+a[1]}9000&&{4d+2a[1]}20000,{d,a[1]}]
Out[]=
{{d1000,a[1]8000}}
In[]:=
Table[a[1]+(-1+n)d,{a[1],{8000}},{d,{1000}},{n,{6}}]
Out[]=
{{{13000}}}
Example 4.3 .8 . The factory “SHUKOM” in the first year of its existence had an income of 6280000 denars, and its costs in the same year amounted to 14280000 denars . With the development of the factory, its costs in the same year decreased every year by 760000 denars, and revenues increased by 840000 denars .
To calculate : a . after how many years the costs will be equal to the revenues and
b . what will be the income and expenses that year?
To calculate : a . after how many years the costs will be equal to the revenues and
b . what will be the income and expenses that year?
In[]:=
Table[a[1]+(-1+n)d[a],{a[1],{6280000}},{d[a],{840000}}]
Out[]=
{{6280000+840000(-1+n)}}
In[]:=
Table[b[1]+(-1+n)d[b],{b[1],{14280000}},{d[b],{-760000}}]
Out[]=
{{14280000-760000(-1+n)}}
In[]:=
Solve[{{6280000+840000(-1+n)}}=={{14280000-760000(-1+n)}}]
Out[]=
{{n6}}
In[]:=
Table[a[1]+(-1+n)d[a],{a[1],{6280000}},{d[a],{840000}},{n,{6}}]
Out[]=
{{{10480000}}}
In[]:=
Table[b[1]+(-1+n)d[b],{b[1],{14280000}},{d[b],{-760000}},{n,{6}}]
Out[]=
{{{10480000}}}
Example 5.3 .3 . ([3]) One machine today is worth € 40.000 . The value of the machine at the end of each month decreases by 6 % of the value that was at the beginning of the month . What will be the value of the machine after two years?
In[]:=
Tablea[1],{a[1],{40000.}},{p,{-6}},{n,{24}}
-1+n
1+
p
100
Out[]=
{{{{9638.3}}}}
Example 5.3 .4 . ([3]) Tourist visit to a tourist place during the period from March 1 to August 31, 2015. constantly growing on the principle of geometric progression . The difference between the visit in June and April is 52.800 visitors, and the difference between May and April of the same year is 24.000 visitors . To calculate : a) what is the percentage of the monthly increase of visitors and how many visitors were in March 2015 and
b) how many visited the tourist place from March 1 to August 31, 2015?
b) how many visited the tourist place from March 1 to August 31, 2015?
In[]:=
Tablea[1],{n,{4}}-Tablea[1],{n,{2}}52800
-1+n
1+
p
100
-1+n
1+
p
100
Out[]=
-1+a[1]+a[1]52800
p
100
3
1+
p
100
In[]:=
Tablea[1],{n,{3}}-Tablea[1],{n,{2}}24000
-1+n
1+
p
100
-1+n
1+
p
100
Out[]=
-1+a[1]+a[1]24000
p
100
2
1+
p
100
In[]:=
Solve-1+a[1]+a[1]52800&&-1+a[1]+a[1]24000
p
100
3
1+
p
100
p
100
2
1+
p
100
Out[]=
{{p20,a[1]100000}}
In[]:=
Table,{a[1],{100000}},{p,{20.}},{n,{6}}
-1+a[1]
n
1+
p
100
-1+1+
p
100
Out[]=
{{{{992992.}}}}
Example 5.3 .5 . ([3]) The factory "SHUKOM" in 1995, produced 360, 000 kg . from a product, and in 2005, 570000 kg . Calculate the average increase in production?
In[]:=
Tablea[1],{n,{1}}360000
-1+n
1+
p
100
Out[]=
{{a[1]}}360000
In[]:=
Tablea[1],{n,{11}}570000
-1+n
1+
p
100
Out[]=
a[1]570000
10
1+
p
100
In[]:=
SolveTablea[1],{a[1],{360000}},{n,{11}}570000.
-1+n
1+
p
100
Out[]=
{{p-204.703},{p-184.706-61.5426},{p-184.706+61.5426},{p-132.355-99.578},{p-132.355+99.578},{p-67.6451-99.578},{p-67.6451+99.578},{p-15.2939-61.5426},{p-15.2939+61.5426},{p4.70254}}
Example 5.3 .6 . ([3]) 1000 tons of a product were produced in January, 1050 tons in February and 1102.5 tons in March . The same ratio of production change is observed in the coming months . To calculate : a) in which month 1551.3 tons were produced, b) production in December and
c) total production for the whole year?
c) total production for the whole year?
In[]:=
Table1+,{a[n],{1551.3}},{a[1],{1000}},{q,{1.05}}
Log
a[n]
a[1]
Log[q]
Out[]=
{{{9.99963}}}
In[]:=
Table[Log[q,{a[n]/a[1]}]+1,{a[n],{1551.3}},{a[1],{1000}},{q,{1.05}}]
Out[]=
{{{{9.99963}}}}
In[]:=
Table[a[1],{a[1],{1000.}},{q,{1.05}},{n,{12}}]
-1+n
q
Out[]=
{{{1710.34}}}
SolveTable,{a[1],{1000.}},{q,{1.05}},{n,{12}}
(-1+)a[1]
n
{q}
-1+{q}
Example 5.3 .9 . In one mine, 6000 tons of ore were mined on the first day, and 12.473, 57 tons on the last day of that time interval . ore and during this time a total of 141.944, 95 tons were excavated . If the production took place on the principle of geometric progression, calculate : a) production growth rate and
b) in how many days was 141.944, 95 tons excavated?
b) in how many days was 141.944, 95 tons excavated?
In[]:=
Table[{a[1]-S[n]}/{a[n]-S[n]},{S[n],{141944.95}},{a[1],{6000}},{a[n],{12473.57}}]
Out[]=
{{{{1.05}}}}
In[]:=
Table[100*{q-1},{q,{1.05}}]
Out[]=
{{5.}}
In[]:=
Table[Log[q,{a[n]/a[1]}]+1,{a[n],{12473.57}},{a[1],{6000}},{q,{1.05}}]
Out[]=
{{{{16.}}}}