These next several problems all relate to this uh scenario where we have an ice cream store and the whatever they say, the is it the profits of the revenue? Probably the revenue given daily sales side of the revenue as a function of temperature T and that's in degrees Fahrenheit and the number of inches of rain and that's an inches, so that's amount of rain. So we can see here that as temperature goes up, profits or sales go up as the rain goes up, sales go down, which basically that's a typically, you know, probably good scenario. I mean halfway decent model, how it's the functional form, I don't know. Um So they asked us the first couple problems is just to evaluate this at a couple points. So they say, Okay, if the temperature is 60°F and we have two inches of rain, so that's a crappy day. Um we get 50 times 20 times three is $3,000 in profit or in in revenue. So that that should probably be on the lower end of things. Well, I mean unless depending on the climate here where the shop is located, I guess if we get down to 40°, we go to zero, we get to five inches of rain, we go to zero. So you know this hopefully this is probably a fairly miserable day for the ice cream shop. So on a good day, what's they were at 80°F and with no rain. And so that gives us 50 times 40 times 10 or $10,000 in sales. So again, that's probably a pretty good day. Now they ask us to take the partial of this with respect to R and then event. So we take the partial respect to our, This is all constant. And then the river of this with respect to our is just -1 So we get -50 Times T -40. And so if we evaluate that at 90° and 1" of rain We get -50 times 50, which is 2500. And what is that? In units? So we have minus $2500 per inch. So at 90° we're losing about um Um $2,500 per inch of rain, additional additional interview rain. Since we're already at one. Well basically for intervene because this isn't a function of our. So at 90° we're losing quite a bit of money if it rains a lot. And then they asked us to take the partial with respect to T. And that's just keeping this in this constant. Again, we just get this driver of this property is one. So and then they ask us to evaluate that again, at this condition with 90° and 1" of rain. And we see that we get $2,000 per degree Fahrenheit. So what's happening here is that every every degree it goes up every degree Fahrenheit. It goes up sell $2000 uh more in ice cream. So again even if there's only even though if there's one inch of rain, so and that this is actually true for no matter what the temperature is. So rain rain is a big effect here um on this and we saw that here. So I mean if you know the rain for for the rate of change of the temperature of the sales for perfectly temperature is a big function of our here. Um And so then they ask us for the the second, so the the S. D. T. D. R. All right, so that means and we do that and we just take we can go back up here and take the derivative of this with respect to T. And we just get minus 50 everywhere. So it's minus $50 per degree Fahrenheit per inch of rain. So what does that mean? Well that means that um well per inch of rain. So if we if we go up 11 F and one inch of rain we're gonna go down $50. So it's basically saying that it's more dependent on rain or inches of rain. That's one inch of rain is a lot of rain. one F isn't really a big deal. So you know it's it's I'm not sure exactly how to interpret that. Um And then the you know partial s partial or partial T. Is just the same thing. Right? So it's minus $50 per degree Fahrenheit per inch of rain. So that's um I guess you can I think that's enough interpretation. Again this is a little hard to really interpret because these different, You know like I said a 1° change in temperature is You know, not much 1" change in rain is a lot.