4.7k views
2 votes
- 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, in dollars, of the dresses made. The owner has
noticed that the maximum daily profit the company has made is
$1600. This occurred when the dresses were sold for $75 each. The
owner also noticed that selling the dresses for $50 resulted in a profit
of $1225. Using a quadratic relation to model this problem, create an
equation for the company's daily profit.

User Mxlhz
by
7.2k points

1 Answer

4 votes
To create a quadratic equation for the company's daily profit based on the selling price of dresses, we can use the general form of a quadratic equation:

\[ p(d) = ad^2 + bd + c \]

Given that the maximum daily profit is $1600 when dresses are sold for $75, we have the point \((75, 1600)\). Plugging these values into the equation:

\[ 1600 = a(75)^2 + b(75) + c \]

Similarly, when dresses are sold for $50, the profit is $1225, giving us the point \((50, 1225)\):

\[ 1225 = a(50)^2 + b(50) + c \]

We now have a system of three equations:

\[ \begin{cases} 1600 = a(75)^2 + b(75) + c \\ 1225 = a(50)^2 + b(50) + c \end{cases} \]

We also know that the quadratic relation has a maximum at the vertex, which is given by the formula \((-b/(2a), f(-b/(2a)))\). In this case, the vertex is \((-b/(2a), f(-b/(2a))) = (75, 1600)\). This provides a third equation:

\[ \frac{-b}{2a} = 75 \]

Now you can solve this system of equations to find the values of \(a\), \(b\), and \(c\), allowing you to formulate the quadratic equation for the company's daily profit.
User Compuguru
by
6.6k points