00:01
We have the function r of x equals 35x minus x squared over x squared plus 35 and that gives the revenue for selling x thousand items.
00:07
We want to find the value of x that maximizes the revenue and we want to find the maximum revenue.
00:13
So, um, if we use the quotient rule to find the derivative of that, so if we have the derivative of a quotient, we're doing the derivative of a quotient u over b, that rule is, um, we would do u prime v minus u v prime all over v squared.
00:38
Okay, so we're going to need that to find the maximum value of this, um, because we want to find a critical point of this function here and that will tell us where this function is maximized.
00:48
So in this case, our u is the numerator of that, of r of x, and that would be 35x minus x squared.
00:55
Our u prime, or the derivative of that, would be 35 minus the derivative of x squared would be 2x.
01:02
And our v is the denominator, which is x squared plus 35, and the derivative of that, the derivative of x squared is 2x, and the derivative of 35 is just zero.
01:12
So we have all four pieces we need now.
01:14
So the derivative of r of x, or r prime of x, would be our u prime times v, so that would be the 35 minus 2x times v, which is x squared plus 35, and then minus, um, u, which is 35x minus x squared times v prime, which is 2x.
01:40
Um, and all that would be over the, um, v, which is x squared plus 35, and that gets squared.
01:54
Okay? so if we simplify this a bit more, that gives us our prime of x to be, if we, uh, multiply out the two binomials in the numerator, that would give us 35x squared plus, um, plus 35 times 35, which is, um, uh, 1 ,225.
02:19
Then negative 2x times x squared would be negative 2x cubed, and then negative 2x times 35 is minus 70x.
02:26
And then we have a negative 1 understood here, and we can multiply that 2x by both of these terms.
02:34
That would give us minus 70x squared, and then plus 2x cubed.
02:44
All that's over x squared plus 35 squared.
02:48
And if we simplify that just a bit further, we can combine our locked terms, uh, the negative 2x cubed and plus 2x cubed cancel, and we're left with 35x squared minus 70x squared is negative 35x squared.
03:01
So we're done with those.
03:02
And we have minus 70x and then plus 1 ,225, and that's over x squared plus 35 squared equals, uh, and we want to set this equal to zero in order to find the critical points.
03:20
So we would have critical points, um, where this derivative is either zero or undefined.
03:27
So it would be undivided if we're dividing by zero.
03:30
So since this value down here, the x squared plus 35 must always be greater than zero, um, because it's impossible to square a number and add and, uh, get a zero or a negative number.
03:43
So it must always be positive and greater than zero.
03:46
So we never have to worry about having an undefined value.
03:49
But this fraction will be zero whenever the numerator is equal to zero.
03:53
So let's set the numerator equal to zero.
03:55
Um, we'll write that down here.
03:58
Negative 35x squared minus 70x plus 1 ,225 equals zero.
04:04
Let's factor out a negative 35.
04:07
That will leave us with negative 35 times x squared plus 2x minus 35 equals zero.
04:14
And let's factor that quadratic trinomial.
04:18
X squared plus 2x minus 35 would factor into x plus seven times x minus five.
04:23
So if we set that equal to zero, we can see the two solutions are x equals negative seven and x equals positive five.
04:31
So, um, we're not, x means how many items are being sold.
04:37
So we're not going to sell a negative amount of items.
04:39
So we would ignore that negative seven for x.
04:41
So x equals five, um, x equals five would maximize the revenue.
04:48
And if we actually want to find the revenue, erase this to get some room, um, to find the actual revenue, we would find r of five, since that's the value of x we want to use to maximize the revenue.
05:09
That would just be, um, 35 times five and then minus, um, five squared.
05:24
All that's over five squared plus, um, 35.
05:30
That would give us, um, 150 over 60, which would simplify to 2 .5.
05:45
So the maximum revenue, the revenue would be maxed when x equals five and the maxed revenue is equal to 2 .5.
05:59
Okay.
06:00
And that may be in thousands of dollars depending on, um, what the revenue was measured in.
06:05
Okay.
06:06
So next we want to find the value of x that will maximize r of x equals c times x minus x squared over x squared plus c for any positive constant c.
06:15
So we may notice that this is the same problem we just did.
06:17
It's, it just replaced the 35 with c.
06:21
So our work's going to be a lot, uh, very similar to what we just did.
06:25
So if we, uh, are using our quotient rule again, uh, u would be the numerator, which is cx minus x squared.
06:35
The derivative of that or u prime, since c is a constant, the derivative of c times x would just be c minus the derivative of x squared would be 2x.
06:43
And then b is the denominator, which is x squared plus c.
06:47
And then the derivative of that or v prime would be the derivative of x squared, which is 2x.
06:52
And since c is a constant, its derivative is just zero.
06:56
So if we use the quotient rule again, we would do our u prime times v.
07:02
So that would be c minus 2x times our v, which is x squared plus c.
07:07
And then we would subtract u times v prime, which is cx minus x squared times 2x...