Respuesta :
Explanation
From the statement, we know that:
0. the company sells 1000 packs of cards per day for $5 per pack,
,1. with every $0.02 reduction in price, 10 more packs a day are sold.
To solve this problem, we define the following variables and functions:
• r = # of price reductions,
• n(r) = # of packs sold by the company as a function of r,
,• p(r) = price per pack (in $) as a function of the # of price reductions,
,• I(r) = income (in $) as a function of the # of price reductions.
Using points 1 and 2, we write the following functions: function of # of packs sold n(r) as:
• the i
[tex]n(r)=1000+10*r,[/tex]• the function of the price per pack p(r):
[tex]p(r)=5-0.02*r,[/tex]• the function for the income I(r) is given by the product of the # of packs sold n(r) and the price per pack p(r):
[tex]I(r)=n(r)*p(r)=(1000+10*r)*(5-0.02*r)=-0.2r^2+30r+5000.[/tex](1) To find the maximum income, we must maximize the function I(r) for r. To do that, we compute and make equal to zero its first derivative:
[tex]I^{\prime}(r)=-0.2*2r+30=0.[/tex]Solving for r, we get:
[tex]\begin{gathered} -0.4r+30=0, \\ 0.4r=30, \\ r=\frac{30}{0.4}=75. \end{gathered}[/tex]We have found that the maximum income is achieved when the # of price reductions is equal to r = 75.
(2) s
[tex]I(75)=-0.2*75^2+30*75+5000=6125.[/tex]We have found that the maximum possible income per day is $6125.
(3) s
[tex]p(r)=5-0.02*75=3.5.[/tex]We have found that the price per pack that maximizes the income is $3.5.
(4) s
[tex]1000*\text{ \$}5=\text{ \$}5000.[/tex]With this new price structure, the company wins $6125 s
Answer• The maximum possible income per day is $6125.
,• The price per pack that maximizes the income is $3.5.
,• The company makes $1125 extra with this new price structure.
