00:01
In this problem, we are given the demand function for a product sold, and we are given its average cost.
00:07
If the production is limited to 100 units, we want to find the profit function in dollars, the number of units that will maximize the profit, and the maximum profit.
00:21
First, let's recall how do we find the profit.
00:24
The profit p of x corresponds to the difference between the revenue with the cost function.
00:35
So we don't yet have the revenue function, but we are given the demand function, and the revenue function will be equal to the demand function times x, giving us 6100x minus one -half x to the power of three.
00:54
Our cost function is already given to us in our problem statements, which means that our profit function p of x will be right to 6100x minus one -half x to the power of three minus 3030 minus 2x, which we can simplify a little bit to 6098x minus one -half x to the power of three minus 3030.
01:36
So this here is our profit function that we want to maximize.
01:48
To maximize our profit function, let's calculate its first derivative.
01:52
Why? because locations for which the first derivative is equal to zero corresponds to critical points, and critical points are generally minima or maxima in our function.
02:03
So let's differentiate.
02:05
P of x, we will obtain 6098 minus three -half x squared...