Question 1186902: The owner of a small clothing company wants to create a mathematical model for the company’s daily profit,
p, in dollars, based on the selling price, d, of the dresses made. The owner has noticed that the maximum daily profit
the company has made is $1200. This occurred when the dresses were sold for $65 each. The owner also noticed that
selling the dresses for $50 resulted in a profit of $900.
a. Using a quadratic relation to model this problem, create an equation for the company’s daily profit.
b. If he set the price to be $60, what would the profit be?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to create a quadratic model for the company's profit and use it to predict profit at a different price:
**a. Creating the Quadratic Model**
Since the maximum profit occurs at a specific price, we can use the vertex form of a quadratic equation:
p = a(d - h)² + k
Where:
* p is the profit
* d is the selling price
* (h, k) is the vertex (the point of maximum or minimum value).
In our case, the maximum profit is $1200 when the price is $65, so the vertex is (65, 1200). This gives us:
p = a(d - 65)² + 1200
Now we need to find the value of 'a'. We can use the other piece of information: when the price is $50, the profit is $900. Plug these values into the equation:
900 = a(50 - 65)² + 1200
900 = a(-15)² + 1200
900 = 225a + 1200
-300 = 225a
a = -300 / 225
a = -4/3
So, the quadratic equation that models the daily profit is:
p = (-4/3)(d - 65)² + 1200
**b. Profit at $60 Price**
Now, we can find the profit when the price is set to $60. Plug d = 60 into the equation:
p = (-4/3)(60 - 65)² + 1200
p = (-4/3)(-5)² + 1200
p = (-4/3)(25) + 1200
p = (-100/3) + 1200
p = -33.33 + 1200 (approximately)
p ≈ 1166.67
Therefore, if the owner sets the price to $60, the estimated profit would be approximately $1166.67.
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