it cost you c dollars each to manufacture and distribute backpacks. if the backpacks sell at x dollars each, the number sold is given by

n=(a/x-c)+b(100-x)

where a and b are certain positive constanst. what selling price will bring a maximum profit?

Profit as a function of x is

P(x) = n (x - c)
= a - cx + 100 bx - bx^2 - cx

Maximum profit occurs when dP/dx = 0
-2c + 100 b = 2xb
x = -c/b + b/50

Check my work.

Well, to figure out the selling price that will bring a maximum profit, we need to work with the given equation. But humor me for a moment, why did the backpack go to therapy? Because it had too many issues to carry!

Alright, let's get back to business. We can start by simplifying the equation:

n = (a/x - c) + b(100 - x)

Expanding it further:

n = (a/x - c) + 100b - bx

Now, we know that profit is equal to the selling price multiplied by the number sold. So we can express profit (P) as:

P = n * x

Substituting the value for n, we have:

P = [(a/x - c) + 100b - bx] * x

Simplifying this further:

P = a - cx + 100bx - bx^2

To maximize profit, we take the derivative and set it equal to zero:

dP/dx = -c + 100b - 2bx = 0

Now, solving for x:

2bx = -c + 100b
x = (-c + 100b) / (2b)

Therefore, the selling price that will bring a maximum profit is (-c + 100b) divided by 2b. But hey, don't forget to consider some other factors like demand, competition, and market conditions!

Remember, laughter is also a great way to maximize happiness, so I hope my attempt at humor also brought a smile to your face.

To find the selling price that will bring a maximum profit, we need to maximize the profit function. The profit (P) can be calculated by subtracting the cost (C) from the total revenue (R):

P = R - C

The total revenue can be calculated by multiplying the selling price (x) by the number of backpacks sold (n):

R = x * n

And the cost can be calculated by multiplying the cost per backpack (c) by the number of backpacks sold (n):

C = c * n

Now let's substitute the formulas for revenue and cost into the profit equation:

P = (x * n) - (c * n)

To simplify the equation, let's substitute the formula for the number of backpacks sold (n) into the profit equation:

n = (a/x - c) + b(100 - x)

P = (x * ((a/x) - c + b(100 - x))) - (c * ((a/x) - c + b(100 - x)))

Simplifying further:

P = a - c*x + b(100x - x^2) - c(a/x - c + b(100 - x))

Now we can expand the equation:

P = a - c*x + 100bx - bx^2 - ca/x + c^2 - cb + cx

Next, let's arrange the terms to simplify the equation:

P = (-bx^2) + (100bx - cx) + (a - ca/x + c^2 - cb - c*x)

Finally, to find the selling price that maximizes profit, we can take the derivative of the profit function with respect to x and set it equal to zero:

dP/dx = -2bx + 100b - c = 0

Solving for x:

-2bx + 100b - c = 0
-2bx = c - 100b
x = (c - 100b) / (-2b)

Therefore, the selling price that will bring a maximum profit is given by (c - 100b)/(-2b).

To find the selling price that will bring maximum profit, we need to maximize the profit function. The profit can be calculated by subtracting the cost from the revenue, which is the selling price multiplied by the number of backpacks sold.

First, let's simplify the expression for the number of backpacks sold (n):

n = (a/x - c) + b(100 - x)

Next, let's express the revenue in terms of the selling price (x) and the number of backpacks sold (n):

Revenue = Selling Price * Number of Backpacks Sold
R = x * n

Since we want to find the maximum profit, we need to find the maximum value of the profit function P, which is given by:

P = R - Cost
P = R - (c * n)

Substituting the expressions for R and n:

P = (x * n) - (c * n)
P = n(x - c)

Now, let's substitute the expression for n from the given equation:

P = [(a/x - c) + b(100 - x)] * (x - c)

Simplifying further:

P = (a - ac/x + b(100 - x) - bc) * (x - c)

Expanding the brackets:

P = (ax - ac + (100b - bx) - bc) * (x - c)
P = ax - ac + 100b - bx - cx + bc

Now, to find the value of x that maximizes the profit, we need to find the critical points of the profit function by taking its derivative with respect to x and setting it equal to zero. Let's differentiate the profit function:

dP/dx = a - b - c

Setting dP/dx equal to zero:

a - b - c = 0

Rearranging the equation:

a - c = b

Therefore, the critical point occurs when a - c is equal to b.

Now, to determine whether this critical point corresponds to a maximum or minimum, we need to analyze the second derivative (d^2P/dx^2). However, since the second derivative is not given in the question, we cannot determine the nature of the critical point.

In conclusion, without additional information, we cannot determine the selling price that will bring maximum profit. We would need the second derivative of the profit function or more explicit constraints on the constants a and b to make a conclusive determination.