00:01
Hi there, so for this problem, we are given some profit function and the profit function is minus x to the three, then this plus nine times x squared, then this plus 120 times x minus 400.
00:20
Once we have this profit function, we're asked to find the maximum profit and the number of units that must be produced and sold to earn that profit.
00:28
So then to maximize this, what we need to do is to derive this with respect to x, the derivative of the first term is minus three times x squared because we take down the exponent and subtract one unit to that exponent, just like this.
00:44
Now, and if we take it down to two, that multiplies the nine, so we will obtain 18 times x plus 120.
00:52
And the derivative of the last term is just zero because it's a derivative of a constant.
00:57
Now we set this equal to zero.
00:59
As you can recognize in here, this is a quadratic equation...