Complete these ordered pairs for this equation. (0, ), (-1, ), (2, ) y=4
1 answer:
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Answer:
a) $3.50
b) 2,500 cups
c) $6,250
Step-by-step explanation:
a) The number of cups of coffee the student sells, n = 2,000 cups
The price at which the student sells the coffee, P = $4.0
The number of extra cups sold for every $0.1 reduction in price = 100 cups
The cost of making one cup of coffee, C = $1.0
Let 'x' represent the set charge for a cup of coffee, we get;
The number of cups, nₙ = 2,000 + 100 × (4.0 - x)/0.1 = 6,000 - 1,000·x
Therefore, the revenue is given as follows;
Revenue, R = x × (6,000 - 1,000·x) = 6000·x - 1,000·x² = 1,000·(6·x - x²)
The cost, C = 1 × 6,000 - 1,000·x = 6,000 - 1,000·x
The profit, P = Revenue - Cost
∴ Profit = 6000·x - 1,000·x² - (6,000 - 1,000·x) = 7,000·x - 1,000·x² - 6,000
Profit, P = 7,000·x - 1,000·x² - 6,000 = 1,000 × (7·x - x² - 6)
At the maximum profit, we have; , we have;
dP/dx = 0
∴ dP/dx = d(1,000 × (7·x - x² - 6))/d
dR/dx = 1,000·d(7·x - x² - 6)/dx = 1,000 × (7 - 2·x) = 0
x = 7/2 = 3.5
The price that gives the maximum profit, x = $3.50
Therefore, the amount the student center should charge for a cup of coffee to maximize profit, x = $3.50
b) We have;
The number of cups, n = 6,000 - 1,000 × 3.5 = 2,500
The number of cups of coffee the student center sells at the price that gives the maximum profit, n = 2,500 cups
c) The maximum profit, = 1,000 × (7 × 3.5 - 3.5² - 6) = 6,250
The maximum profit, = $6,250.