First, the revenue that the company makes is equal to the camera's price per unit times the number of units sold, or Rev = x*p = x(400-.04x). You need to find determine this function's derivative and find the value for x where the derivative is zero. That will tell you how many cameras they should sell to either maximize or minimize revenue; to determine which it is, you can take the 2nd derivative. If the 2nd derivative at the point x is negative then you have a maximum; if it's positive then you have a minimum. Once you have the value of x for the maximum figured out, you can then plug that back into the equation for p to determine the selling price.
Profit is equal to revenue minus costs, or in this case x*p-(2000+160x). Again, you can find the value of x that maximizes profit by taking the derivative and setting it to zero. |