let the number of $0.50 decreases be n
number of ornaments sold = 120+20n
selling price of each = $4.00 - $0.5n
revenue = (120+20n)(4-.5n)
= 480 - 60n + 80n - 10n^2
= 480 + 20n - 10n^2
If you know Calculus
d(revenue) = 20 - 20n = 0
n = 1
price should be 4-.5 = $3.50
By completing the square:
revenue = -10(n^2 - 2n + 1 - 1) + 480
= -10((n-1)^2 - 1) + 480
= -10(n-1)^2 + 490
max revenue is $490 when n = 1
or the price is 4-.5 = $350
An artisan can sell 120 garden ornaments per week at $4 per ornament. For each $0.50 decrease in price, he can sell 20 more ornaments.
a) determine algebraic expressions for the price of a garden ornament and the number of ornaments sold.
b) write an equation for the revenue using your expressions from part a).
c) use your equation from part b) to find what price the artisan should charge to maximize revenue.
1 answer
1 answer