it costs 8 dollars to manufacture and distribute a backpack.If the backpack sell at x dollar each, the number sold , n, is given by n= 3/(x-8)+5(100-x) . Find the selling price that will maximize the profit.
profit=n(x-8)
Profit= 3/(x-8)+5(100-x) * (x-8)
= 3+5(100-x)(x-8)
dProfit/dx=0= 500 (x-8)+5(100-x)
0=500x-5x-4000+500 check that.
x= 3500/495 check all that, I did it in my head.
Why did the backpack want to maximize its profit? Because it wanted to make sure it was carrying the most "change" possible! 🎒💰
To find the selling price that will maximize the profit, we need to first find the profit equation. The profit from selling n backpacks at x dollars each is given by profit = revenue - cost. The revenue is the total amount received from selling the backpacks, which is n * x, and the cost is the total manufacturing and distribution cost, which is 8 * n.
So, the profit equation is:
profit = n * x - 8 * n
Now, we'll substitute the given equation n = 3/(x-8) + 5(100-x) into the profit equation:
profit = (3/(x-8) + 5(100-x)) * x - 8 * (3/(x-8) + 5(100-x))
Let the calculator handle all that tedious math. Let's find the selling price, x, that maximizes the profit! 🎉
To find the selling price that will maximize the profit, we need to determine the profit equation and then find the value of x that maximizes it.
The profit equation is given by: Profit = Revenue - Cost.
The cost of manufacturing and distributing a backpack is $8.
So, the cost equation is: Cost = $8.
The revenue equation is given by: Revenue = Selling price * Number sold.
Given: n = 3/(x-8) + 5(100-x), where n represents the number sold.
Substituting the value of n into the revenue equation:
Revenue = x * (3/(x-8) + 5(100-x)).
Simplifying:
Revenue = x * (3/(x-8) + 500-5x).
Revenue = x * (502 - 4x) / (x-8).
Now we can evaluate the profit equation:
Profit = Revenue - Cost.
Profit = x * (502 - 4x) / (x-8) - $8.
To maximize the profit, we need to find the critical points by taking the derivative of the profit equation and setting it equal to zero:
d(Profit) / dx = 0.
Let's find the derivative of the profit equation:
d(Profit) / dx = [(x-8) * (4x - 502) - x * (-4)] / (x-8)^2.
d(Profit) / dx = (4x^2 - 504x + 4016 + 4x) / (x-8)^2.
d(Profit) / dx = (4x^2 - 500x + 4016) / (x-8)^2.
Setting the derivative equal to zero:
(4x^2 - 500x + 4016) / (x-8)^2 = 0.
To solve this equation, we can factorize the numerator:
4(x^2 - 125x + 1004) = 0.
We can then solve for x using the quadratic formula:
x = (-(-125) ± sqrt((-125)^2 - 4(1)(1004))) / (2(1)).
x = (125 ± sqrt(15625 - 4016)) / 2.
x = (125 ± sqrt(11609)) / 2.
x = (125 ± 107.75) / 2.
So the possible values for x are:
x1 = (125 + 107.75) / 2 = 232.75 / 2 = 116.375.
x2 = (125 - 107.75) / 2 = 17.25 / 2 = 8.625.
However, x = 8.625 is not a valid solution since it would result in a denominator of 0 in the number sold equation.
Therefore, the selling price that will maximize the profit is $116.375.
To find the selling price that will maximize the profit, we first need to determine the profit equation. The profit can be calculated by subtracting the cost (manufacturing and distribution) from the total revenue.
The total revenue is given by the selling price multiplied by the number sold (n). Therefore, the revenue equation can be written as:
Revenue = Selling Price * Number Sold
Since we are given the equation for the number sold, we can substitute it into the revenue equation:
Revenue = Selling Price * (3/(x-8) + 5(100-x))
Now, to find the profit, we subtract the cost from the revenue:
Profit = Revenue - Cost
= Selling Price * (3/(x-8) + 5(100-x)) - 8
To maximize the profit, we need to find the value of x that maximizes the profit equation. This can be done by taking the derivative of the profit equation with respect to x and setting it equal to zero:
d(Profit)/dx = 0
Let's differentiate the profit equation:
d(Profit)/dx = d/dx (Selling Price * (3/(x-8) + 5(100-x)) - 8)
To differentiate the equation, we can use the product rule and chain rule:
d(Profit)/dx = Selling Price * (d/dx (3/(x-8))) + d/dx (5(100-x)) - 0
Now, let's calculate the derivatives:
The derivative of 3/(x-8) with respect to x can be found by using the chain rule:
d/dx (3/(x-8)) = -3/(x-8)^2
The derivative of 5(100-x) with respect to x is simply -5, since 100-x does not contain x.
Now, we can substitute these derivatives back into the equation:
d(Profit)/dx = Selling Price * (-3/(x-8)^2) - 5
Setting this equation to zero, we can solve for x:
0 = Selling Price * (-3/(x-8)^2) - 5
Simplifying, we get:
5 = -3/(x-8)^2
Cross-multiplying, we obtain:
(x-8)^2 = -3/5
Since the square of a real number cannot be negative, this equation has no real solutions. It means that there is no maximum profit value in this scenario.
In conclusion, according to the given equation, there is no selling price that will maximize the profit.