How to solve difficult algebra problem?
A College Bookstore has been selling one tee shirt for a year. During Fall term, 100 sold at $5 each. During Spring term, price raised to $7; consequently, 80 sold
a) Suppose number of tee shirts sold (N) depends linearly on price charged (x).Write equation showing this dependence. Identify and explain meaning of slope & vertical intercept. Then graph function
b) According to this relationship, how many tees would be sold during a term at $3.50 each? What should cost of a shirt be to sell 45 per term?
c) Suppose it costs Bookstore $3 per shirt
• Calculate profit during terms where they charge: $5,$6,$7,$8,$9,$10
• Use data to explain why profit doesn't depend linearly on price
• Explain why profit, P(x) depends quadratically on price
• Determine parameters a, b and c in quadratic model P(x)=ax^2+bx+c
d) Use different way to determine parameters a, b and c in quadratic model p(x)=ax^2+bx+c. Here is how: Generate formula describing how term profit (P) depends on price charged (x).HINT: PROFIT = (NUMBER SOLD) * (PROFIT PER SHIRT)
e) Draw graph of quadratic model, P(x).Use graph to find P-intercept; what does this point represent?
f) Calculate break-even points (price to charge in order to yield profit of $0).Explain meaning of each break-even point in context of situation
g) Now use graph to find price that will maximize Bookstore's profit.What will that profit be?
Thank you for any possible help!!
Comments
Let N=ax+b, where a, b are constants.
(5,100)=>
100=5a+b------(1)
(7, 80)=>
80=7a+b-------(2)
From (1) & (2) get
a=-10, b=150
=>
(a) N=-10x+150
slope=-10=>
for the increase of $1, there are
reduction of 10 tee shirts in sale;
the N-intercept=150 meaning that
150 shirts will be "sold at no charge".
(b) N(3.5)=-10(3.5)+150=115 sold.
N=45=>x=(150-45)/10=$10.5/shirt
(c) N=150-10x; let $C=the cost/shirt;
the profit P=$(X-C)N
| X |..5..|..6..|..7..|..8..|..9..|..10..|
| C |-------------3---------------------|
| N |100|.90.|.80.|.70.|.60.|.50...|
| P |200|270|320|350|360|350..|
By inspecting the data, P(x) does
not linearly depend on x.
By regression, get
P(x)=-10x^2+180.0039x-449.9219 approximately
[SSE=200.75, s^2=66.91666]
=>a=-10, b=180, c=449.9219
By solving equations, taking 3 points:
(10,350), (8,350) & (6,270) from the table.
P(x)=ax^2+bx+c
=>
100a+10b+c=350
64a+8b+c=350
36a+6b+c=270
=>
a=-10,b=180,c=-450
=>
p(x)=-10x^2+180x-450
=>the profit depends quadratically on the charge x.
p(0)=-450 meaning that the Bookstore will lose
$450 totally in profit if it "sells" tee shirts at no charge.
For maximizing P, let
P'=-20x+180=0=>x=9
P(9)=max. profit=$360
=> If the store sells tee shirt at $9 each, it will
get a max.profit of $360 in total.
Points given are (5,100) and (7,80)
Slope = -20/2 = -10
Equation
y-100 = -10(x-5)
y = -10x+150