Question 1132937: a florist sells 17000 bouquets per year at a cost of 9 dollars each. In the previous year they raised the price to 16 dollars, they only sold 10000 bouquets that year. Assuming the amount of bouquets sold is in a linear relationship with the cost, what price should the florist charge per bouquet to maximize their revenue?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! when the price is 9, the quantity is 17000
when the price is 16, the quantity is 10000
when the price goes up 7, the quantity goes down 7000.
7000 / 7 = 1000
for every unit that the price goes up,the quantity goes down 1000 units.
this is what they mean when they say the amount of bouquets sold is in a linear relationship with the cost.
that particular linear relationship can be shown as a straight line on a graph.
you have two possible equations.
the first one is where x is the number of units sold and y is the price.
the second one is where x is the price and y is the number of units sold.
here's what the equation and graph of the first one looks like.
y represents the price and x represents the number of bouquets sold.
here's what the equation and graph of the second one looks like.
y represents the number of bouquets sold and x represents the price.
the revenue equation, however, is not a straight line.
the equation is revenue = price * quantity.
when the price is 9, the quantity is 17000 and the equation becomes revenue = 9 * 17000.
you can solve for revenue to get 153000 = 9 * 17000.
153000 is the revenue when the price is 9 and the quantity sold is 17000.
from the linear relationship between price and quantity, you can see that when the price goes up one unit, the quantity goes down 1000 units.
letting x be the unit, then the formula becomes:
y = (9 + x) * (17000 - 1000 * x)
that's a quadratic equation.
apply the distributive law to get:
y = (9 + x) * (17000 - 1000x) becomes y = 9 * 17000 - 9 * 1000x + x * 17000 - x * 1000x
combine like terms to get y = 153000 + 8000 * x - 1000 * x^2.
order the terms in descending order of degree to get y = -1000x^2 + 8000x + 153000.
in this format, you get:
a = coefficient of x^2 term = -1000
b = coefficient of x term = 8000
c = constant term = 153000
y will be at its maximum value when x = -b/(2a) = -8000 / -2000 = 4.
when x = 4, y will be equal to -1000 * 16 + 8000 * 4 + 153000 = 169000.
that will be the maximum revenue.
the graph of the equation (9 + x) * (17000 - 1000x) and the equation -1000x^2 + 8000x + 153000 will be identical as shown on the following graph.
the maximum point on the graph is when x = 4.
when x = 4, the price is 9 + 4 = 13.
when x = 4, the quantity is 17000 - 4000 = 13000.
revenue = price * quantity = 13 * 13000 = 169000.
the coordinate point of (4,169000) on the graph shows that.
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