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A. What is the expected weekly demand for the alarmclock radio?The expected weekly demand is(Type an integer or a decimal. Do not round.)b. What is the probability ...

Question

A. What is the expected weekly demand for the alarmclock radio?The expected weekly demand is(Type an integer or a decimal. Do not round.)b. What is the probability that weekly demand will be greaterthan the number of available radios?The probability is(Type an integer or a decimal. Do not round.)c. What is the expected weekly profit from the sale of the alarmclock radio? (Remember: There are onlyfourclock radios available in any week to meet demand.)The expected weekly profit is(Round to the nea

a. What is the expected weekly demand for the alarm clock radio? The expected weekly demand is (Type an integer or a decimal. Do not round.) b. What is the probability that weekly demand will be greater than the number of available radios? The probability is (Type an integer or a decimal. Do not round.) c. What is the expected weekly profit from the sale of the alarm clock radio? (Remember: There are only four clock radios available in any week to meet demand.) The expected weekly profit is (Round to the nearest cent as needed.) d. On average, how much profit is lost each week because the radio is not available when demanded? The expected weekly profit lost is (Round to the nearest cent as needed.) I need the steps to the solutions for each question.



Answers

Business The monthly revenue $R$ achieved by selling $x$ wristwatches is figured to be $R(x)=75 x-0.2 x^{2}$ $x$. Wristwatches is figured to be $R(x)=75 x-0.2 x^{2}$ The monthly cost $C$ of selling $x$ wristwatches is $$ C(x)=32 x+1750 $$ (a) How many wristwatches must the firm sell to maximize revenue? What is the maximum revenue? (b) Profit is given as $P(x)=R(x)-C(x) .$ What is the profit function? (c) How many wristwatches must the firm sell to maximize profit? What is the maximum profit? (d) Provide a reasonable explanation as to why the answers found in parts (a) and (c) differ. Explain why a quadratic function is a reasonable model for revenue.

So the first part asks how many, uh, MP three players will be supplied at a market price of $80 So we just substitute to in these up by equation and sulfur X. So I have 88 minus 25 divided by 10.5, and this gives me six million units supplied. I can do the same to find the number of units demanded. That's idiot. Minus 1 40 divided by negative 5.2. That's 10 million units demanded Andi. Its supply less than demand. Yes, it is in part B, we do the same, but with a different price. I have 114. So at this price, this apply this eight point for a 1,000,000,000 units and for demand I have five million units. Is demand less than supply? Yes, it is on Park Si asks. At what price does the market reach equilibrium? So I just equate both the equations. Yeah. 10.5 X plus 25 calls 1 40 minus 5.2 x Just give me 15. 27 X equals 115 and therefore X is just 11 If I divide by 15.7 and that's 7.32 million units. From this, I can find the equilibrium price, so that's 10.5 times 7.3 to plus 25 which is $101 on 86 cents. So that's what it's gonna be an equilibrium.

Okay, So we'll be looking at some problems that combine both linear and quadratic functions, particularly with respect to manufacturing. And the 1st, 1st example we're gonna look at has to do with uh, wristwatches. So a company is selling making and selling wristwatches and there's what's called a revenue function. That just means the amount of money that they make on the rift, not profit, but the amount of money that they bring in the income relative to the risk, the wrist watches. And that revenue function is 75 x minus 2/10 X squared. And then we also have a cost function that is the cost to make the wrist watches. And so when we look at those two graphs together, then the the if we're looking first of all, let's just look just at the revenue function. So if we want to find out how many wristwatches we need to sell in order to maximize revenue without any regard to cost, then we're gonna maximize revenue right here At 187.5. So obviously we can't sell half a wristwatch. So our revenue would be maximized at either 187 wristwatches or 188 wristwatches. Both of those would give us maximum revenue. So it's gonna max out at 187 or 188 revenue maxes out At 187 Or 1 88. And that maximum revenue would be whatever value we get when we substitute 187 or 188 into The function. So it would be, we're gonna take 75 multiply that by 187 -2/10 Times 187. And that gives us a maximum revenue of $13,987 and 60 cents. Oh no, that's not right, I didn't square that. It's 75 times 187 minus 2/10 Times 187 Squared. Here we go. That's better. Um and that's a maximum revenue of $7,031.20. I knew that it wasn't right because it didn't agree with. There wasn't close to that Vertex maximum there. So when we sell 187 or 188 Wrist watches, we have an income, just an income, not a profit of an income of $7.31, 20 cents. Now, the profit function, if you think about this realistically a profit function is just gonna be the cost that it takes to the revenue, the total that you bring in for the sale of whatever product or selling minus the cost to make that product and that's going to be called the revenue. The profit function. So the profit function is going to be simply the revenue, the amount that you bring in minus whatever it costs you to make that product. So when we substitute These in, we get a profit function of 75 x minus two tents, x squared- your cost function of 32 x plus 1750. And then when you distribute that negative sign, you get a profit function, you distribute the negative sign and you re arrange your variables to make to write your expression in standard form, you get negative 2/10 X squared plus 43 X -1 750. So that's going to be your profit function. And then we'll look at our profit function in order to maximize profit. So then when we graph the profit function, it's gonna look like this here, this purple graph And clearly we're gonna have a maximized profit here at 107 five wristwatches. So profit is maximized on either side of that profit is maxed At 107 or 108 watches. And that maximum profit Is going to be. So we're just gonna take p of 107 and that gives us negative point to times 17 squared Plus 43 times 1 7 -1750. And that gives us a profit of $561.20. Now notice it's not exactly 561 25 because that's if we were just at 107 and a half watches, which we clearly can't do. Um so your maximum profit is also going to be the same at 108 because remember that axis of symmetry of your parabola goes straight down the middle. So these two points on either side of the vertex are going to be um reflections of each other across that That axis of symmetry. So your profit is maximized at 107 or 108 watches for a maximum profit of $560.20. And then if you think about it, um why is the profit max? Why are these two answers different? So why is the revenue max? And the prophet max is different? Well, clearly it's because the revenue max doesn't take into account the cost to produce the material or to produce the product. In this case the wristwatch.

Okay, So we'll be looking at some problems that combine both linear and quadratic functions, particularly with respect to manufacturing. And the 1st, 1st example we're gonna look at has to do with uh, wristwatches. So a company is selling making and selling wristwatches and there's what's called a revenue function. That just means the amount of money that they make on the rift, not profit, but the amount of money that they bring in the income relative to the risk, the wrist watches. And that revenue function is 75 x minus 2/10 X squared. And then we also have a cost function that is the cost to make the wrist watches. And so when we look at those two graphs together, then the the if we're looking first of all, let's just look just at the revenue function. So if we want to find out how many wristwatches we need to sell in order to maximize revenue without any regard to cost, then we're gonna maximize revenue right here At 187.5. So obviously we can't sell half a wristwatch. So our revenue would be maximized at either 187 wristwatches or 188 wristwatches. Both of those would give us maximum revenue. So it's gonna max out at 187 Or 188 revenue maxes out At 187 Or 1 88. And that maximum revenue would be whatever value we get when we substitute 187 or 188 into The function. So it would be, we're gonna take 75 multiply that by 187 -2 Tents Times 187. And that gives us a maximum revenue of $13,987 and 60 cents. Oh no, that's not right. I didn't square that. It's 75 times 187 minus 2/10 Times 187 Squared. Here we go, That's better. Um and that's a maximum revenue of $7,031.20. I knew that it wasn't right because it didn't agree with. There wasn't close to that Vertex maximum there. So when we sell 187 or 188 Wrist watches, we have an income, just an income, not a profit of an income of $7.31, 20 cents. Now, the profit function, if you think about this realistically a profit function is just gonna be the cost that it takes to the revenue, the total that you bring in for the sale of whatever product or selling minus the cost to make that product and that's going to be called the revenue. The profit function. So the profit function is going to be simply the revenue, the amount that you bring in minus whatever it costs you to make that product. So when we substitute These in, we get a profit function of 75 x two tents, x squared- your cost function of 32 x plus 1750. And then when you distribute that negative sign, you get a profit function of you distribute the negative sign and you re arrange your variables to make to write your expression in standard form, you get negative 2/10 X squared plus 43 X -1750. So that's going to be your profit function. And then we'll look at our profit function in order to maximize profit. So then when we graph the profit function, it's gonna look like this here, this purple graph. And clearly we're going to have a maximized profit here at 107 .5 Wristwatches, so profit is maximized on either side of that profit is maxed At 107 or 108 watches. And that maximum profit Is going to be, so we're just gonna take p of 17 and that gives us negative point to times 17 squared Plus 43 times 1 7 -1 750. And that gives us a profit of $561.20. Now notice it's not exactly 561 25 because that's if we were just at 107 and a half watches, which we clearly can't do. Um so your maximum profit is also going to be the same at 108 because you remember that axis of symmetry of your parabola goes straight down the middle. So these two points on either side of the vertex are going to be um reflections of each other across that That axis of symmetry. So your profit is maximized at 107 or 108 watches for a maximum profit of $560.20. And then finally why do the two answers in A. And C. Differ wise, revenue different than profit. And clearly revenue is different than profit because the revenue doesn't include the cost to produce. The revenue just includes the amount of money taken in from the sale of the profits from the sale of the product, whereas the profit includes the cost to produce as well. And then why is a quadratic function? A reasonable model for revenue? A quadratic function is a reasonable model for revenue because revenue is equal to price times quantity. So the price of the item comes, the quantity of items that you sell. The price is going to be a function of the number of wristwatches sold. And so the revenue is the product of two functions of X. Um, so when you have, it's two functions of X, so it's gonna be a quadratic.

For this problem, we are told that a company manufactures and sells X. Smartphones per week. The weekly price, demand and cost equations are p equals 500 minus 0.5 X. And c. F X equals 20,000 plus 1 35 X. For the first part, we are asked what price should the company charged for the phones and how many phones should be produced to maximize the weekly revenue? And what is the maximum weekly revenue then for party? We are asked what is the maximum weekly profit? And how much should the company charge for the phones? How many phones should be produced to realize the maximum weekly profit keep in mind profit versus revenue. So from part A regarding the price, uh and specifically the weekly revenue, We know that the revenue is going to equal the price, times the number of units sold. So our revenue is going to actually equal X times 500 -0.5 x 0.5 X. Which then means we can expand this out as negative 0.5 X squared Plus 500 x one second here. Now to maximize the weekly revenue, we want to take the derivative of the revenue with respect to the number of units sold. So that is going to give us, so we'll have negative 0.5 times two. So that's going to be negative one X plus 500. So we want to figure out how many units sold will bring this to zero. Obviously that is going to be just when X equals 500. We'll have that the The derivative of the revenue with respect to units sold is going to be zero. And we can tell, you know, this is a parabola. So up here is going to be the maximum, that's the turning point where the derivative zero. So to figure out what the maximum weekly revenue then will be, we just plug that back in. So we'll have our equals 500 times 500 -1 half of 500. So it's going to be 500 times 2 50 one second here. So this will be $125,000 weekly revenue. Then for part B were asked about the maximum weekly profit. So the profit is going to be the revenue minus the cost. So that will be for us. We had that, the revenue was negative 0.5 X squared plus 500 X. And then when we subtract off the cost, uh we'll have 500 minus 1 35 X. So that is going to be one second here. So we'll have negative 0.5 X squared plus 3 65 x minus 20,000. So to maximize the profit, we take the derivative with respect to units sold again. That will give us negative X Plus 3 65. And we want to solve this for zero. So we'll just get that X needs to equal 365. So the maximum profit we'll be when we plug that in, so I'm going to pause. All right. So the maximum profit will be 46,612.5.


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