SOLUTION: callenge:hard A shoe maker sells a new model of athletic shoe for $90. On average, the store sells 45 pairs of these shoes each week. Jason, the sale manager, estimates that he

Question 935446: callenge:hard
A shoe maker sells a new model of athletic shoe for $90. On average, the store
sells 45 pairs of these shoes each week. Jason, the sale manager, estimates that
he will sell 4 additional pairs of shoes each week for every $5 reduction in the
price .Find the following
1- A quadratic function modeling the store�s average weekly revenue from the
sale of these shoes as related to reduction in price.
2- The range of price which would allow he store to maintain or increase its
current weekly revenue from these shoes.
3- The maximum amount by which the store can increase its revenue from these
shoes each week . Round of to the nearest whole dollar.

Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Let n = the integer number of $5 reduction in price of the athletic shoes.
Let I = the total weekly income (revenue).
We have an income of the number of shoes sold times the price of the shoes
(1) I = (45 + 4n)*(90 - 5n)
Simplifying (1) we get
(2) I = 45*90 - 45*5n + 4*90n -4*5n^2 or
(3) I = 4050 - 225n + 360n - 20n^2 or
(4) I = 4050 + 135n - 20n^2 or
(5) I = - 20n^2 + 135n + 4050
Equation (5) is the quadratic solution asked for in part 1.
The answer to part 2 is that value of n at which the value of I goes below 4050. We need to set (5) equal to zero and find the two values of n that make (5) equal to 4050. We can see that I of (5) is equal to 4050 when do not change the price, i.e. n = 0. At n=0 (5) becomes
(6) -20*0 + 135*0 + 4050 = 4050
The quadratic of (5) increases as n increases, however when the first two term sum to a negative value we get less than 4050. This happens when
(7) -20n^2 + 135n = 0 or
(8) n*(-20n + 135) = 0 or
(9) n = {0,6.75)
Answer to part 2; The price range is (90, 90-5*6.75) or ($90,$56.25)
Part 3 asks for the maximum value of I given by (5). This occurs when the derivative of (5) equals zero (if you're into calculus) or the apex of the parabola for algebra students. The answer is the same
(10) Imax occurs at n = 135/(2*20) or
(11) n = 3.375
Put this value of n into (5) to get
(12) Imax = -20n^2 + 135n + 4050 or
(13) Imax = -20(3.375)^2 + 135*3.375 + 4050 or
(14) Imax = -20n^2 + 135n + 4050 or
(15) Imax = -227.8125 + 455.625 + 4050 or
(16) Imax = 227.8125 + 4050 or
(17) Imax = 4277.8125
Answer to part 3; The maximum weekly revenue is $4278.

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