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A retailer has been selling 1200 tablet computers a week at $ \$ 350 $ each. The marketing department estimates that an additional 80 tablets will sell each week for every $ \$ 10 $ that the price is lowered.(a) Find the demand function.(b) What should the price be set at in order to maximize revenue?(c) If the retailer's weekly cost function is $$ C(x) = 35,000 + 120x $$what price should it choose in order to maximize its profit?

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04:58

Wen Zheng

01:55

Carson Merrill

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Julia R.

May 12, 2020

Missouri State University

University of Nottingham

Idaho State University

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Hi there for a problem that we need to hold too late First the demand function. And we are given two points that 1000 in 200 tablets, computers. Our sell by $350. And also that if we and sell um an additional of 80 tablets. So this will be dance. We can sell that for $10 less. So that will be $340. So this is the given information and we need to find the demand function. So ah we will set that P. Is the price and Q. Is the number of tablets sold. So that the function of price should be something like this. Where m. Is the slope of dysfunction. And P zero corresponds to a given point. And And Q002 another given point for the number of top of that let sold. So to find first the slope we know that that is change and in price over the change in the number of chocolate salt. So we can see in here that we put the final price by the initial price over. Then find a number minus the initial number. So the final price or the second point in here is given by this because the p um the B component or the price is greater. That so that will be the final the final part. Oh no sorry. This is the price. This is the christ. This is the initial um number of tablets. Uh huh. This is the final one. This is the initial price and this is the final price. So with that set which just simply substitute those. So it will be 340 -350 over mm So this will give us -1/8. And this is the slope. So we just substitute this slope in this equation and we put them some values for piece of C. Rank is zero. I'm going to use this pointing here So we will have pete minus piece of zero which is 350 is equal to the slow times cute minus 1000 and 200. Yeah. So solving for P we will have that function is 1/8 you loss 500. And this is the demand function. No for part B of this problem. And is what should the price be said in order to maximize the revenue. So in this case we need to find the revenue function. Um They divided fined the maximum and then substitute that value in the price function. That's what we need to do for power bi. So the revenue function. Okay that will come art of Q. Is the price function or the demand function times the number of have let sold. So that if we multiply that function bike you will have 1/8. Q square plus by 100 Q. And to find a critical point. We I don't know if it is a minimum or a maximum for to find a critical point. We divide this function with respect to Q. So we will have 1/4 cube glass. 500. And to find a critical point. We said this to see rope. So it is very straightforward that will obtain Q. is equal to 200. So this is our a critical point. So to know if this critical point is a maximum or a minimum, we need to do A second derivative of the revenue function with respect to queue. So that will obtain -1/4. And this value is less than zero. So that we will have a maximum At you over 200. And with that we just think this is to do that value in the function of the price. So it will be and is functioning here. They were obtained for the previous eaten. So we will have 1 1/8. Two trends blast Okay and if we block this into the calculator we obtain $250. No for parse theme of this problem we need to oh um we are given some cost function and we are asked that what price should it choose in order to maximize its profit. So function that we are given is Yeah that's and in this case as represent the number of of tablets sold. So it is equal to Q. And to obtain the profit. The function of profit is equal to the revenue minus the cast function. So from the previous we'll have the revenue function and so that will be that the prophet function of ads in this case we'll kind of do it with ads but we you can do it also with Q. It doesn't matter. So it will be one over eight. That's a squirt plus 500 adds minus the gas function that we are given. So it will be mine is 35,000. Uh huh. I'm sorry this is -2 because all of the cows function um should be multiplied by the miners. So this will be test um We can simplify tests. We will have one over feet squared. Um This will be to us 380 ask mind is this so again we want to maximize this function and then with that value that we obtained we substituted in the function for the price. Um so that we will maximize the profit. So again to find a critical point of the function we in the rebate with respect to the variable X. In this case we will have 1/4 at last. 380. Okay. And we set this to zero the hood because we want a critical point so that we will find that adds yes 1000 and fight 120 to know if this value is um is a maximum or a minimum. We need to revive these two times and again we'll obtain that this is minus 1/4. This value is less than zero. So we will have a maximum But at equals two the value that we just have obtained. And we just simply need this institute that value in the price function. So the price function remember is the first one that week obtained this in here. Q. Is equal to add. Remember that? So dot We will have the price that makes the maximize that profit is equal to minus 1/8. The value that we obtain for ads which is this um blast by hundreds and this give us $310. So these And or charging $310 per tablet will not see mind the profit. So thank you so much and this is it for this problem.

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