at $1.00 per expresso
income = 200*1 = $200
cost = 200 + 1*200 = $400
at $2.00 per expresso
income = 100*2 = 200
cost = 200 + 100 = 300
This question makes absolutely no sense to me.
What am I missing here ?
This is a losing enterprise!
Last time expesso cost $1.00 was in the 1970's
To produce espressos, a coffee shop has fixed costs of 200 dollars each day and variable costs of one dollar per espresso. The number of espressos that the coffee shop sells on a given day depends linearly on the price of each espresso: If the price is $1.00, then they sell 200 espressos, and if the price is $2.00, then they sell 100 espressos. What is the choice of the price that will maximize their profit?
2 answers
if they they sell x espressos for price p each, x = ap+b
a+b = 200
2a+b = 100
a = -100 and b = 300
so x = -100p + 300
revenue for x espressos is xp = -100p^2 + 300p
cost is 200+x = 200 + -100p+300 = -100p+500
profit is thus (-100p^2+300p)-(-100p+500) = -100p^2 + 400p - 500
Did I miss something? This function is always negative, so they never make a profit. Just as a sanity check, we know that:
if the price is $1.00, they sell 200, for revenues of $200 and costs of $300
if the price is $2.00, they sell 100, for revenue of $200 and costs of $400
??? ... ???
a+b = 200
2a+b = 100
a = -100 and b = 300
so x = -100p + 300
revenue for x espressos is xp = -100p^2 + 300p
cost is 200+x = 200 + -100p+300 = -100p+500
profit is thus (-100p^2+300p)-(-100p+500) = -100p^2 + 400p - 500
Did I miss something? This function is always negative, so they never make a profit. Just as a sanity check, we know that:
if the price is $1.00, they sell 200, for revenues of $200 and costs of $300
if the price is $2.00, they sell 100, for revenue of $200 and costs of $400
??? ... ???
0 answers