Question 1110688: To maximize yearly revenue, a company wishes to determine the best-selling price for their new line of shirts. They realize that if the price is too high or too low, they will not maximize profit. The equation y=-16(x-30)^2+10 000 models the scenario, where x represents the selling price per shirt and y represents the number of shirts sold
Determine the selling price that maximizes the revenue
Determine the maximum number of shirts sold
What is the maximum revenue
What are the x-intercepts for the function?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the equation is y = -16 * (x-30)^2 + 10,000
simplify to get y = -16 * (x^2 - 60x + 900) + 10,000
simplify further to get y = -16x^2 + 960x - 14,400 + 10,000
combine like terms to get y = -16x^2 + 960x - 4,400
set y = 0 to get -16x^2 + 960x - 4,400 = 0
since this is in standard quadratic form, then:
a = coefficient of x^2 term = -16
b = coefficient of x term = 960
c = constant term = -4,400
the maximum value is when x = -b/2a.
solve for x to get x = -960 / -32 = 30
when x = 30, the original equation becomes y = -16 * (30-30) + 10,000
solve for y to get y = 10,000
looks like the maximum number of shirts to be sold is 10,000 when the price per shirt is equal to 30.
here's the graph of the original equation.
the maximum revenue will be 10,000 * 30 = 300,000.
the x intercepts for the function will be when the value of y is equal to 0.
the original equation becomes -16(x-30)^2 + 10,000 = 0
subtract 10,000 from both sides to get -16(x-30)^2 = -10,000
divide both sides by -16 to get (x-30)^2 = -10,000 / -16
take the square root of both sides to get x-30 = plus or minus sqrt(10,000/16)
simplify to get x = 30 = plus or minus 25.
solve for x to get x = 5 or 55.
y = 0 when x = 5 or when x = 55.
the graph shows that as well.
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