A meat department manager discovrs that she can sell m(x) killograms of ground beef in a week, where m(x) = 14 700 - 3040x, if she sells it at x dollars per kilogram. She pays her supplier $3.21/kg for the beef.
a) Determine an algebraic expression for P(x), where P(x) represents the total profit in dollars for 1 week.
b) Find the equation for the inverse relation.
c) Write an expression in function notation to represent the price that will earn $1900 in profit. Evaluate and explain.
d) Determine the price that will maximize profit.
e) The supply cost drops to $3.10/kg. What price should the manager set? How much profit will be earned at this price?
a) To determine the total profit in dollars for 1 week, we need to subtract the cost of purchasing the beef from the revenue earned by selling it. The cost of purchasing the beef is the quantity sold (m(x)) multiplied by the cost per kilogram ($3.21). The revenue earned is the quantity sold (m(x)) multiplied by the selling price per kilogram (x). Therefore, the algebraic expression for P(x) is:
P(x) = m(x) * x - m(x) * $3.21
b) To find the equation for the inverse relation, we need to interchange the roles of x and m(x) in the original equation. Let y represent m(x) and x represent the selling price per kilogram. Then we have:
x = 14,700 - 3,040y
Solving this equation for y will give us the inverse relation.
c) To represent the price that will earn $1900 in profit, we need to solve the equation P(x) = $1900 for x. This can be done by substituting the expression for P(x) derived in part a) and solving the resulting equation. After solving for x, we will have the price per kilogram that will earn $1900 in profit.
d) To determine the price that will maximize profit, we need to find the maximum value of the P(x) function. This can be done by finding the critical points of P(x), i.e., the values of x where the derivative of P(x) is zero. We can then use these critical points to find the maximum value of P(x).
e) To determine the price the manager should set when the supply cost drops to $3.10/kg, we can repeat the steps from part c) using the updated cost per kilogram. This will give us the price per kilogram that should be set. Additionally, we can calculate the profit earned at this price by substituting the value of x into the P(x) function derived in part a). This will allow us to determine the profit earned at the new price.
cost c(x) = 3.21*m(x)
revenue r(x) = x*m(x)
profit p(x) = r(x)-c(x)
Now go for it.