Kaylee’s Quality Quadratic Coffee Company (KQQCC) sells tall lattés for $3.25 ea
ID: 376770 • Letter: K
Question
Kaylee’s Quality Quadratic Coffee Company (KQQCC) sells tall lattés for $3.25 each. Assume this is the only product KQQCC sells. KQQCC has fixed costs of $1,200 per month, and its actual cost to produce each latté is 95 cents.
A. KQQCC sells 2,000 lattés per month at the price of $3.25. Kaylee experimented with raising the price to $3.75 and found that sales dropped to 1,600 lattés per month. Assuming that this relationship between price and sales is linear, find a formula for the number of lattés she sells as a function of the price that she charges. Then find a formula for her revenue as a function of the price she charges, and determine the price she should charge, rounded to the nearest five cents, to maximize revenue.
B. KQQCC’s profit is its revenue minus its total costs. Determine the price it should charge to maximize profit.
Explanation / Answer
A.
Y = lattes
X = price
linear equation: Y = mX + c
when latte = 2000, price = 3.25
hence: 2000 = 3.25*m + c.............equation 1
when price = 3.75
latte = 1600
hence, 1600 = 3.75m + c......... equation 2
solving equaltion 1 &2, subtracting them gives us
2000 - 1600 = 3.25m - 3.75m + c - c
400 = -0.5m
m = -800,
putting value to equaltion 2000 = 3.25*-800 + c, c = 4600
demand functoin of latte: Y = -800X + 4600
Revenue = Price*quantity sold
= X*(-800X + 4600)
Revenue = -800X2 + 4600X
Maximum revenue when the derivative with respect to price of revenue function is equal to 0.
-1600X + 4600 = 0
X = 2.875
Hence, to maximize revenue, she should charge $2.875
B.
Profit = revenue - cost
cost = fixed + variable
fixed cost = 1200
variable cost per latte = 95 cents = 0.95
total variable cost = 0.95*quantity sold
= 0.95*(-800X + 4600)
= -760X + 4370
total cost = 1200 -760X + 4370
= -760X + 5570
Profit = -800X2 + 4600X - (-760X + 5570)
= - 800X2 + 5360X - 5570
Again to maximize profit the derivative with respect to price of profit function is equal to 0.
-1600X + 5360 = 0
X = $3.35
Hence, she should charge $3.35 to maximize profit.
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