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The answers to most of the following exercises are in terms of logarithms and exponentials. A calculator can be helpful, enabling you to express the answers in decimal form.To encourage buyers to place 100 -unit orders, your firm's sales department applies a continuous discount that makes the unit price a function $p(x)$ of the number of units $x$ ordered. The discount decreases the price at the rate of $\$ 0.01$ per unit ordered. The price per unit for a 100 -unit order is $p(100)=\$ 20.09$.a. Find $p(x)$ by solving the following initial value problem.$$\begin{aligned}&\text { Differential equation: } \quad \frac{d p}{d x}=-\frac{1}{100} p\\&\text { Initial condition: } \quad p(100)=20.09\end{aligned}$$b. Find the unit price $p(10)$ for a 10 -unit order and the unit price $p(90)$ for a 90 -unit order.c. The sales department has asked you to find out if it is discounting so much that the firm's revenue, $r(x)=x \cdot p(x),$ will actually be less for a 100 -unit order than, say, for a 90 -unit order. Reassure them by showing that $r$ has its maximum value at $x=100$.d. Graph the revenue function $r(x)=x p(x)$ for $0 \leq x \leq 200$.

a.) $p(x) = 20.09 e^{1 - \frac{x}{100}} $ b). $ p(10) \approx \$49.15 $ and $ p(90) \approx \$22.20 $ c.) Find derivative of $R(x)$ and should with $1^{st}$ derivative test that $ x= 100$ have a max. d.) See video for graph.

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

Campbell University

University of Nottingham

Idaho State University

Boston College

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So we're told that this differential equation here talks about a discount that is applied to our function P with X number of units sold. So they first asked us to solve for this equation to find what P. T. Is. So let's do that. And we actually don't even need toe do any calculus for this because they tell us in this chapter, when we have a differential equation of the form off, the derivative is equal to k. Times are function, then it should have a solution of p of tea is equal to peanut or about P. T. P of X in this case, since that's our variable and it will just be peanut our initial amount times E to the K X in this case. So it would just be and this right here is K, so it would be X over negative 100. So that's going to be our solution for that differential equation. Now we can use our initial condition here to solve for peanuts, so we have p of 100 is equal to 20.9 and this is P not times e to the negative. Well, it is 100 over 100. So that just simplifies down toe one so we'd have negative one. So to get peanut, we would just want to multiply each side by e. So that's going to give. Peanut is equal to 20 times 0.9 e. All right that says we get P of X is equal to 20.9 e times e to the negative x over 100 which we can simplify down to just 20 0.9 e to the one minus X over 100. So this will be our solution or a differential equation. Now they want to find p of 10 as well as P of 90. All right, so we can just plug those in so p of 10 should be 20 0.9 times e to the one minus 10 over 100. So that should be 90 over 100 in the numerator, which would give us approximately, Let's see, so hee raise to the 90 over 100. There's about 2.4. Multiply that by 20.9 and that should give us about 49 dollars and 15 cents now. What about P of 90. Well, it's gonna be 20.9 times E to the one minus 90 over 100 which is going to be approximately. So he raised to the so one minus 90 or 90 over 100 which would just be 9/10 should be 1/10. So it's e to the 1/10 which is about 1.1 multiplied by 20.9 And that gives us a price of about $22.20. So these air going to be the price for selling each of the 10 units and selling each of the 90 units. Now they tell us that someone at the company's having issues with thinking. We're selling everything for two less. We're not making any money. So we want to now show Is that a revenue function, which is going to be X times p of X So the number of units times our price has a maximum at 100 units sold? All right, well, let's go ahead and write out what p of X are. Our revenue function is first, so that's going to be 20.9 times X and then e to the one minus X over 100. All right, now, to find maximums, remember, we look at the derivative first derivative, so our prime of X is equal to so we'll need to use chain rule to take this derivative so that constant we could just move out. So the first terms derivative ex would just be one side B e to the one minus X over 100. They were gonna have plus x times. The derivative art next function which eats the X derivative, is just itself. But then we apply chain rules. So we take the derivative of its exponents which should end up being negative won over 100. All right, now, we want to set this equal 20 and doing that, we can divide by our constant here. And we can also divide by this e to the one minus X over 100 cause that can never be equal to zero, which would imply that one minus 100 over X is equal to zero. Well, let me write that a little bit better, and then solving for X, we would get X is equal to 100 and then we would use wth e first derivative test to show that it is a max. So I'll use first derivative test to check. So that's how we would actually quell their, uh, concerns about this. Now they want us to Raph, this function on zero toe 200. So let's go ahead and do that. So our revenue function again. Is this up here? So first you might notice when X is equal to zero. When we sold no units, we should have no revenue. So we're going to start Bear and let's look at the limit as this goes to infinity. So the limit as X goes to infinity is going to be Well, we know that e to the negative ax should dominate act. So that means our vic should go to zero. So we know eventually it should flatten out. And well, then we also need to find our max so we can put that down So we don't want to look at our of 100 which is going to be 20.9 times 100. Let me zoom out a little times 100 and then e to the one minus 100 over 100 would be e to the zero, So this would just be our revenue. So actually, I didn't need to zoom out for that. And that should be 2000 and $9. So let's just put that appears 2000 and nine, and that's going to be at 100. And so we have that point and the next thing we'd want to do. So let's find X is equal to 200 as well because we need to plot that, cause at least you know, it should look something kind of like this. Then it's going to start coming down. Since that start maximum and then we would just have whatever 200 it is. So are of 200 is going to be actually do this in a different color to make it stand out a little bit more. Who so our ah 200 is going to be 20.9 times 200 then we'd have e to the so it be one minus 200 over 100 which would be one minus two so e to the negative one, which is approximately equal to it looks like $1478.14. So something around 14 78 and so this point here at 200 to be 14 78. So that's just a decent sketch of our graph. Or you can just plug it into a calculator toe, make it look even better. But I think at least for doing it by hand, this is about as good as we really want.

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