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?
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
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
??? ... ???
To find the price that will maximize the coffee shop's profit, we need to consider the relationship between the price, the number of espressos sold, and the costs incurred.
Let's start by analyzing the relationship between the price and the number of espressos sold. We are given that the number of espressos sold depends linearly on the price. Specifically, if the price is $1.00, they sell 200 espressos, and if the price is $2.00, they sell 100 espressos. This information suggests that the relationship between the price (P) and the number of espressos sold (N) can be represented by a linear equation of the form N = mP + b.
To find the slope (m) and the y-intercept (b) of this linear equation, we can use the given data points. Let's plug in the first data point, where the price is $1.00 and they sell 200 espressos:
200 = m(1) + b
Next, let's plug in the second data point, where the price is $2.00 and they sell 100 espressos:
100 = m(2) + b
Now we have a system of two equations with two variables (m and b). We can solve this system of equations to find the values of m and b.
Using the first equation, we can rewrite it as:
m + b = 200 (equation 1)
Using the second equation, we can rewrite it as:
2m + b = 100 (equation 2)
To solve this system of equations, we can subtract equation 1 from equation 2:
(2m + b) - (m + b) = 100 - 200
Simplifying, we get:
m = -100
Now we can substitute this value of m into equation 1 to solve for b:
(-100) + b = 200
b = 300
Therefore, the linear equation that represents the relationship between the price (P) and the number of espressos sold (N) is N = -100P + 300.
Now let's move on to calculating the coffee shop's profit at each price point. Profit can be calculated as revenue minus costs. Revenue is equal to the price (P) multiplied by the number of espressos sold (N), and costs are equal to the fixed costs plus the variable costs per espresso multiplied by the number of espressos sold.
We are given that the fixed costs are $200 per day, and the variable costs are $1 per espresso. Therefore, the total costs can be calculated as:
Total Costs = Fixed Costs + Variable Costs per Espresso * Number of Espressos Sold
= $200 + $1 * N
= $200 + $1(N)
= $200 + N
The revenue can be calculated as:
Revenue = Price * Number of Espressos Sold
= P * N
Finally, the profit can be calculated as:
Profit = Revenue - Total Costs
= P * N - ($200 + N)
= PN - $200 - N
Now that we have the profit equation, we can substitute the linear equation for N into the profit equation:
Profit = P * (-100P + 300) - $200 - (-100P + 300)
= -100P^2 + 300P - $200 + 100P - $300
= -100P^2 + 400P - $500
To find the price that maximizes the profit, we can take the derivative of the profit equation with respect to P, set it equal to zero, and solve for P.
Taking the derivative of the profit equation:
d(Profit)/dP = -200P + 400
Setting the derivative equal to zero and solving for P:
-200P + 400 = 0
-200P = -400
P = 2
Therefore, the choice of price that will maximize the coffee shop's profit is $2.00 per espresso.