I've been trying to work this problem out for the past week and still havent managed to get to the answer.
jeans unlimited sold 6000 pairs of jeans in october at an average of 44$ each. The company president decides that she must increase the company's profits by increasing the price of a pair of jeans. Market research indicates that sales will drop 200 pairs for every 1$ increase in price. If the cost of making a pair of jeans is 25$ and all jeans made are sold , what price will maximize the profit?
The answer is 49.50$ but im not sure how to get there. I know that i somehow have to find a quadratic equation and find the derivative of that to find the max which will essentially give me enough information to find the final answer. PLEASE HELP.
Profit=NumberJeans*price-cost
but numberjeans=6000-200(Price-44)
profit=(6000-200(P-44))P-25
I assume you know how to find max.
dp/dPrice=0=(6000-200(P-44))+P(-200)(1)
solve for P
0=6000-200P+8800-200P
400P=14,800
P= 47 dollars. Check my work.
I recommend on your graphing calculator graph the profit function
Profit vs P
To find the price that maximizes the profit, we can start by calculating the profit function.
Let's break down the given information:
- Jeans sold in October: 6000 pairs
- Average price per pair: $44
- Cost of making a pair: $25
- Sales drop per $1 price increase: 200 pairs
First, let's determine the revenue function, which is the product of the number of jeans sold and the price per pair. In this case, it is given by:
Revenue = (Number of jeans sold) x (Price per pair)
Revenue = 6000 x $44 = $264,000
Next, we need to consider how the change in price affects the number of jeans sold. Since sales drop by 200 pairs for every $1 increase in price, we can write the equation for the number of jeans sold as:
Number of jeans sold = (Starting number of jeans sold) - (Change in price in dollars * Sales drop per dollar increase)
Number of jeans sold = 6000 - (Price increase in dollars * 200)
We want to find the price that maximizes profit, which is given by the equation:
Profit = (Revenue - Cost) = (Number of jeans sold) x (Price per pair) - (Number of jeans sold) x (Cost per pair)
Profit = (6000 - (Price increase in dollars * 200)) x (Price per pair) - (6000 - (Price increase in dollars * 200)) x (Cost per pair)
Now, we can simplify the profit equation:
Profit = (6000 - 200x) * (44x) - (6000 - 200x) * 25
Profit = (264000 - 8800x) - (150000 + 5000x)
Profit = 264000 - 8800x - 150000 - 5000x
Profit = -13800x + 114000
To find the price that maximizes profit, we need to find the value of x (price increase in dollars) that maximizes the profit function. In this case, x represents the number of dollars by which the price is increased.
To maximize the profit, we can take the derivative of the profit function with respect to x and set it equal to 0 to find the critical point. Then, we can determine if it is a maximum or minimum by checking the second derivative or by analyzing the behavior of the function.
Differentiating the profit function:
d(Profit)/dx = -13800
Setting the derivative equal to 0 to find the critical point:
-13800 = 0
Since -13800 is a constant, it is never equal to 0. Therefore, this equation has no critical points.
Since the profit function is a linear equation (-13800x + 114000), this means that the function increases or decreases linearly without any maximum or minimum.
However, the problem states that we must increase the price to increase profits. Therefore, we need to analyze the behavior of the profit function as x increases.
If we increase x (price increase in dollars), the profit function (-13800x + 114000) decreases linearly. Since we want to maximize the profit, we need to find the highest possible value for x.
Considering the given information, as the price increases, sales drop by 200 pairs for every $1 increase in price. This implies that if we increase the price too much, the decrease in sales will offset the increase in revenue, resulting in lower profit.
To maximize the profit, we need to find the highest possible value for x that does not result in a decrease in profit. In this case, we need to find the maximum price increase that does not decrease the quantity sold below 0.
Setting the quantity sold equal to 0 to find the maximum price increase:
6000 - 200x = 0
6000 = 200x
x = 6000/200
x = 30
Therefore, the maximum price increase is $30.
Now, let's calculate the maximum price by adding the price increase to the original price per pair:
Max price = $44 + $30
Max price = $74
Hence, the price that maximizes the profit is $74 per pair of jeans.
Apologies, there seems to be a discrepancy between the given answer and the process described above. Please recheck the problem statement or consult the correct solution for further clarification.