00:01
Okay, so we're given that as you can advance those 500 cones daily when charging two pounds per cone and that they find out when they increase the amount by one euro that they sell 100 less cones.
00:15
So when to first find the demand equation, so the demand equation is going to be like a linear function.
00:24
So like p of x is going to equal mx plus b.
00:29
Well, we're given that our slope is.
00:33
That there is a for every one euro increase there's a 100 less cones right instead of losing so therefore a slope is one over 100 negative one over 100 for every one dollar euro increase we lose 100 cones or we sell 100 less cones so now if they give our information we need to just find b so we can use the the fact that at two dollars or two euros we sell 500 cones uses to find our intercept so this is 2 is equal to negative 5 plus b so we get that b is equivalent to 7 so our demand equation would be p of x equal to negative 1 over 100 x plus 7 so now we want to find the daily revenue equation so revenue is just our unit times the price.
01:56
So we already know our price equation, which is negative 100x plus 7.
02:00
So we need to multiply that by x.
02:02
So we should end up with negative x squared over 100 plus 7x.
02:13
Now, part c, we want to find the price that should be charged to maximize the daily revenue.
02:19
So to do so, we can go a few ways.
02:22
We can take the derivative.
02:22
Or we can use the equation x it was negative b over 2a.
02:28
So let's just do the derivative in this case.
02:31
So let's find the derivatives of our revenue equation.
02:37
That's what by doing so, we end up with negative x over 50 and plus 7.
02:45
And doing this, and then the derivative, the derivative approach, we want to set this equal to zero...