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8) A manufacturer determined that its marginal cost per unit produced is given by tefncio Ck) 0.0006x2 _ 0.4x + 76. Find the total cost of producing the 31st unit t...

Question

8) A manufacturer determined that its marginal cost per unit produced is given by tefncio Ck) 0.0006x2 _ 0.4x + 76. Find the total cost of producing the 31st unit through the 40Oth unit

8) A manufacturer determined that its marginal cost per unit produced is given by tefncio Ck) 0.0006x2 _ 0.4x + 76. Find the total cost of producing the 31st unit through the 40Oth unit



Answers

Find the marginal cost for producing units. (The cost is measured in dollars.) $$ C=205,000+9800 x $$

So to find the marginal cost function from our cost function, All we have to do is we need to find the first derivative of our cost function, since the marginal cost is equal to the first derivative of our cost function or C. Prime. So we can say m is equal to the derivative with respect to X. Of 4500 plus 1.47 And so we can do a couple things to this derivative. First, if we have a derivative of one term plus another term, we can split that up so it's equal to the derivative of the first term plus the derivative of the second term. And now we can do is we can first find this derivative of 4500, which is just the derivative of a constant, which means it's equal to zero. So this is equal to zero plus the derivative of 1.47 times X. The next thing we can do is we can take out this 1.47 since it's a constant and we can multiply it by the derivative of just X. So this is equal to 1.47 times the derivative of X. And now to find the derivative of X. All we have to do is use the power rule, and so this is equal to 1.47 times one, which is equal to 1.47 So are marginal cost function. M is equal to 1.47

All right, so this is problem. Three of calculus Early Transcendental Chapter eight section for So the question is the marginal cost of producing X units of a certain product is 74 close 1.1 ax minus 0.2 x squared plus 0.4 x cubed, so in dollars per unit. So we want to find the increase in cost if the production level is raised from 1200 to 1600 units. So we essentially want to find a delta in the cost function going from 1200 units to 1600 units. This is not limit notation or anything. It's just saying the change in cost when you go from producing 1200 units to 1600 and that's gonna just be equal to C minus. See the cost. The cost of 1600 units producing 1600 units minus the cost of producing 1200 units, which if you look at this, you'll see we have a function evaluated at one place, minus the same function evaluated and another which should remind you of this statement of what an integral is. So if we have and to go from a to b some function we'll call f DF dx DF the integral, the integral from A to B of the derivative of X. With respect to acts, D x is just equal to f of be minus. Oh, yet that's what I want minus f uh, A So that's look similar. So what we need on the other side of our cost equation is the change in cost with respect to units produced. And we need to integrate that. And as we see here, the marginal cost of X units is equal to the derivative of cost with respect to the amount of units. So we're gonna plug in here since they gave us the marginal cost. Ray, we're gonna plug that in to marginal cost, which gives us 74 plus 1.1 ax minus 0.2 There's a point there X squared plus and we're just going to do Ah, we're after right for zeros X cubed. And that equals D C. Dia. So the change in costumes, respect acts is the marginal cost. So now we want to solve for the cost function at 1600 minus the cost function and 1200. And when we see from our integral weaken, get that just by integrating with bounds. So we're going to do that. We're gonna integrate both sides of the equation from 1200 to 1600 and obviously this right side is gonna be trivial. I'm gonna kind of box stuff up so that nothing gets confusing about what's anywhere. So the right side of this equation integrating the costs, that derivative of costumes, respect toe acts. And I forgot already I needed d x here. Same with here, which is not gonna fit in well, but you need to integrate with respect to something, and in this case, it's axe DX, and obviously the integral of a derivative with respect to X DX is just going to give us the cost function. But we have our bounds. So we need to plug in, are bound, so see, of 1600 minus sea of 1200 see of 1200. And then now we have to do the left side, which is where the problem really lies, which is integrating this and then evaluating it. So we're gonna first just integrate and leave our evaluation to the next step. So integrate were just right now bringing our bounds down because we're only integrated now, integrating a polynomial we're just going to use the power rule the reverse power rule which says that the integral of some of the integral of X to the a d X equals X to the A plus one divided by a plus one. So obviously I'll give a real example right here. So the integral inter girls also, you can separate this integral by addition. So we don't need to worry about addition. We can just do it one term of the polynomial at a time. Integration. So we have 74 and we want to raise the power of X from zero to a one and divide by the power raised to so we can put a divided by one there. But obviously divided by one does nothing. And now, for the next one ex now has power one. So we need to We need to divide by the power we raised two. And since we're going from X to the 1st 2 x two the second we need to divide by two. But she see there and raised the power, obviously, and for the next one Same were always just gonna bring our co efficient, which is 0.2 And now we're raising X from 2 to 3. So x to the second two x 1/3. So we divide by three and our last term is again Just bring the coefficient, which is 0.4 That's a four. And we want to raise the power from X cubed toe X to the fourth and divide by the power he raised to. And then this is all evaluated at 1600 to 1200 which is what we're gonna do next. And we're also so we're gonna say dealt Ah, see, 60 out 1200 to 1600. The change in cost going from producing 1200 units to 1600. Because we just replaced this. I went to that. We went to Delta Psi just for notation. And now we need to plug in, which means we're gonna start with we're gonna start with plugging in the upper bound, so plugging in for acts 1600 here 74 times 1600 and it's to the first power. Plus you're not try slanted plus Now we plug in for X in the second term, 1.1 divided by two. And then now we have 1600 plug done but squared. And then minus 0.2 divided by three and then plug in 1600. And this time it is cubed. And then plus our last term here, zero 0.4 zeros four divided by four times X to the fourth, and we're plugging in 1600 for X 1600 X is raised to the fourth. So to the fourth. And then now we're gonna plug in 1200. And from our equation for the definition of integration with bounds, which is boxed in right here, we need to now have a minus and evaluate the function. Serena? Uh, yeah, area. So we're just gonna wrap this in parentheses to make it more clear to read. Then we're gonna have ar minus and evaluate at the lower bound of the into girl, which is 1200 in this case. So now we just right almost the same thing. And I'm gonna go to a new line here because I don't have room or is on Tilly, but we have evaluate at 1600 minus. Now we need to evaluate it. 1200 which is gonna be the same statement. But with 1200 in there instead of 1600 1.1, divided by two, 1200 squared. My zero point you're a zero to divided by three time 1600 cubed and then plus last term for zeros. None of four. Bye bye. Four. And then now we have. Oh, and I made a mistake here. I hope you caught me, but that's not 1600 were evaluating at the lower Bound, which is 1200. And then we just need our last 1200 to the fourth. So now we've got our whole expression for the change in cost from 1200 to 1600. There's no variables in there, which means we can evaluate it to a number. So, um, actually, I since I already have it plugged in here, I'm gonna save you the time of watching me plug this into my calculator. But there's our answer, which is 20,464,000 and 800 dollars change. All right, so that's the problem. And I guess actually I'll just state it again. Delta PSI 1200 to 1600 Change and cost Going from producing 1200 to 1600 units is equal to what we found, which is 20,464,800. So that's a

All right, so this is problem. Three of calculus Early Transcendental Chapter eight section for So the question is the marginal cost of producing X units of a certain product is 74 close 1.1 ax minus 0.2 x squared plus 0.4 x cubed, so in dollars per unit. So we want to find the increase in cost if the production level is raised from 1200 to 1600 units. So we essentially want to find a delta in the cost function going from 1200 units to 1600 units. This is not limit notation or anything. It's just saying the change in cost when you go from producing 1200 units to 1600 and that's gonna just be equal to C minus. See the cost. The cost of 1600 units producing 1600 units minus the cost of producing 1200 units, which if you look at this, you'll see we have a function evaluated at one place, minus the same function evaluated and another which should remind you of this statement of what an integral is. So if we have and to go from a to b some function we'll call f DF dx DF the integral, the integral from A to B of the derivative of X. With respect to acts, D x is just equal to f of be minus. Oh, yet that's what I want minus f uh, A So that's look similar. So what we need on the other side of our cost equation is the change in cost with respect to units produced. And we need to integrate that. And as we see here, the marginal cost of X units is equal to the derivative of cost with respect to the amount of units. So we're gonna plug in here since they gave us the marginal cost. Ray, we're gonna plug that in to marginal cost, which gives us 74 plus 1.1 ax minus 0.2 There's a point there X squared plus and we're just going to do Ah, we're after right for zeros X cubed. And that equals D C. Dia. So the change in costumes, respect acts is the marginal cost. So now we want to solve for the cost function at 1600 minus the cost function and 1200. And when we see from our integral weaken, get that just by integrating with bounds. So we're going to do that. We're gonna integrate both sides of the equation from 1200 to 1600 and obviously this right side is gonna be trivial. I'm gonna kind of box stuff up so that nothing gets confusing about what's anywhere. So the right side of this equation integrating the costs, that derivative of costumes, respect toe acts. And I forgot already I needed d x here. Same with here, which is not gonna fit in well, but you need to integrate with respect to something, and in this case, it's axe DX, and obviously the integral of a derivative with respect to X DX is just going to give us the cost function. But we have our bounds. So we need to plug in, are bound, so see, of 1600 minus sea of 1200 see of 1200. And then now we have to do the left side, which is where the problem really lies, which is integrating this and then evaluating it. So we're gonna first just integrate and leave our evaluation to the next step. So integrate were just right now bringing our bounds down because we're only integrated now, integrating a polynomial we're just going to use the power rule the reverse power rule which says that the integral of some of the integral of X to the a d X equals X to the A plus one divided by a plus one. So obviously I'll give a real example right here. So the integral inter girls also, you can separate this integral by addition. So we don't need to worry about addition. We can just do it one term of the polynomial at a time. Integration. So we have 74 and we want to raise the power of X from zero to a one and divide by the power raised to so we can put a divided by one there. But obviously divided by one does nothing. And now, for the next one ex now has power one. So we need to We need to divide by the power we raised two. And since we're going from X to the 1st 2 x two the second we need to divide by two. But she see there and raised the power, obviously, and for the next one Same were always just gonna bring our co efficient, which is 0.2 And now we're raising X from 2 to 3. So x to the second two x 1/3. So we divide by three and our last term is again Just bring the coefficient, which is 0.4 That's a four. And we want to raise the power from X cubed toe X to the fourth and divide by the power he raised to. And then this is all evaluated at 1600 to 1200 which is what we're gonna do next. And we're also so we're gonna say dealt Ah, see, 60 out 1200 to 1600. The change in cost going from producing 1200 units to 1600. Because we just replaced this. I went to that. We went to Delta Psi just for notation. And now we need to plug in, which means we're gonna start with we're gonna start with plugging in the upper bound, so plugging in for acts 1600 here 74 times 1600 and it's to the first power. Plus you're not try slanted plus Now we plug in for X in the second term, 1.1 divided by two. And then now we have 1600 plug done but squared. And then minus 0.2 divided by three and then plug in 1600. And this time it is cubed. And then plus our last term here, zero 0.4 zeros four divided by four times X to the fourth, and we're plugging in 1600 for X 1600 X is raised to the fourth. So to the fourth. And then now we're gonna plug in 1200. And from our equation for the definition of integration with bounds, which is boxed in right here, we need to now have a minus and evaluate the function. Serena? Uh, yeah, area. So we're just gonna wrap this in parentheses to make it more clear to read. Then we're gonna have ar minus and evaluate at the lower bound of the into girl, which is 1200 in this case. So now we just right almost the same thing. And I'm gonna go to a new line here because I don't have room or is on Tilly, but we have evaluate at 1600 minus. Now we need to evaluate it. 1200 which is gonna be the same statement. But with 1200 in there instead of 1600 1.1, divided by two, 1200 squared. My zero point you're a zero to divided by three time 1600 cubed and then plus last term for zeros. None of four. Bye bye. Four. And then now we have. Oh, and I made a mistake here. I hope you caught me, but that's not 1600 were evaluating at the lower Bound, which is 1200. And then we just need our last 1200 to the fourth. So now we've got our whole expression for the change in cost from 1200 to 1600. There's no variables in there, which means we can evaluate it to a number. So, um, actually, I since I already have it plugged in here, I'm gonna save you the time of watching me plug this into my calculator. But there's our answer, which is 20,464,000 and 800 dollars change. All right, so that's the problem. And I guess actually I'll just state it again. Delta PSI 1200 to 1600 Change and cost Going from producing 1200 to 1600 units is equal to what we found, which is 20,464,800. So that's a

All right, so calculus were given a marginal cost function. In this example, the marginal cost function is 74 plus 1.1 x minus 0.2 x squared plus 0.4 X cubed. So, um, we want to find what is the difference in cost of production when you go from producing 1200 units to 1600 units and the units are just how maney items are you producing And then different to the units were talking about our units as in dollars dollars car item produced, which are the units of our function. So we want to find what is the change in the price of producing the whole batch of items when you go from producing 1200 items to 1600 items and we are given a marginal cost function, which I just read. So we have this relation, which is that the marginal cost of an item is equal to the change in the total cost with respect to the amount of items, I, which is they're the derivative of the total cost with respect to the amount of items. So the first thing we're gonna do is just take this formula and replace. We're gonna replace this with our actual function. So to do that, we obviously displayed in here. So 74 plus 1.1 ax minus 0.2 x squared plus 0.4 x cubed equals. And then we didn't change anything about the right side of the equation or we didn't substitute anything on the right side. So we just have the change in the derivative of constant respect toe acts. And if you remember, if you kind of define a derivative so it's a the integral from our defining into go the into girl from A to B of the derivative of some function af with respect acts, um, de X is just gonna be equal to f of be minus f of a. And that's exactly what we want. We want something in the form like this where we have a function evaluated at B minus of function, evaluated somewhere else because that's what our answer is gonna look like here. We have the cost function at 1600 minus the cost function at 1200 so we can see to get to the ghetto, having this format of ah function evaluated at a value minus the same function evaluated a value we're gonna have to integrate with bounds, that air the values we want. And we you can see here this is just a derivative just like this. So to go to, to go, to having the actual function, see, instead of its derivative and with plugged in numbers in the format that we're trying to figure out right here, we are going to need to what we did in this formula which has taken derivative or taken integral with correct bounds with respect to the r d with a with a, um, integrating piece that is the same as the derivative is with respect to. So just clear that up. And we're gonna just do that We're gonna right into girl. And in this case are bounds are 1200 because we want we want 1200 to be in the, um, subtracting position right here. So we're gonna put a to 1200 and then we're gonna put B to 1600 B to 1600 and a is 1200 so that means 12 hundreds gonna go here and 1600 is going to go at the top so that we get what we want once we evaluate the end zero and you can't just take an integral on one side without taking an integral on the other. So on this side, we're also going to write down. And I'm gonna make a little room there, gonna write down and take the entire girl on this side too, and with the same bounds. Because if you don't do the same thing to both sides, you won't maintain equality. And this is the integral not just of 74 but of the whole function. So now obviously this integral. We just used to get the right side and to grow, which is use this formula, which is going to give us the cost of 1600 items produced minus the cost of 1200 items, which is exactly what we're trying to solve for right is we're trying to find that change in the cost from going from producing 1200 items to 1600 items and then the other side is not as simple, but we just need toe integrate. Oh, and I forgot my DX is here is important to have those because they do matter. So we're not gonna We're just gonna integrate the statement and leave, um, plugging in the bounds to afterwards. I'm gonna I like this too, so that it's obvious to see so toe integrate a Paulino meal. We just use the power rule on each term and the power rule. The power rule just states that for an integral of the integral of X to the A equals X to the A plus one where a is a constant divided by a plus one and into girls you can separate by addition, so we can do it for each each, um, polynomial term of x each term of the polynomial individually. So that just gives us 74 the power of X zero on 74. So we raised the power toe one and divide by one for 1.1 X. Now the power of X is one, and we need to raise it to two. And we also need to divide the coefficient by two. So 1.1 divided by two times X and were raised, the power two squared and then minus 0.2 because we're just 002 because we're just gonna bring the coefficients down. And this time we're raising the power 23 So we need to divide by three and then our last term again, Same thing. And we're just gonna bring our co efficient and then raise the power. In this case, it's from 3 to 4 and then divide by the power of raised to which is for so and then the last thing we need, which I'm gonna need to erase to make room for you. Have tow. Forgive me for that. But we need to remember to keep our evaluation because all we did was integrate what's inside. We didn't evaluate at our bounds and in this case are bounds are still 1200 to 1600 for the specific problem and then just rewrite what we had before, which was the change in cost going from 1600 to 1200. So now we just need to evaluate are integral. And that is a rough 1600 right there, Some in a right over that 1600. And so we want to find this are right side is all good. This is what we want, But we want the left side to be a number because it's asking for what is the difference in price. It's not asking for a nun evaluated statement. So to evaluate those brackets is pretty simple. And this side, we're just gonna right Delta Psi from 1216 100 Just another way of writing it. The change in cost going from 1200 to 1600 items produced. And so to evaluate the integral we just basically we're gonna have here minus and here. Plus, and we're gonna plug in inside the plus side, we're gonna plug in the upper bound, which is 1600. So it's gonna be 74. Time 1600 plus one point 1/2 time 1600 its X squared. So 1600 squared minus 0.2 divided by three. And and I raced to make room probably should write it left to right. And now we have x cubed and we're plugging in 1600 again. So it's gonna give us 16 on dread cubed. And then our last term, which is plus and then four zeroes 0.0 for do I buy for times 1600 and now we have X to the fourth. So we I'm gonna put to the fourth there, and then I'm going to go to a new line. But then, just like in here, let me do some highlighting. So that blue part corresponds to here in our definition of an integral Now we're going to do the red part, which is the minus. So we're gonna have minus and then here I'm gonna move to a new line, just Teoh, keep space on the white board, not squeeze everything in, and we're just gonna evaluate the same function. But instead we're gonna have s of a which is the lower bound of the integral, which is 1200 so will have almost the exact same thing. So 74 times this time our input is 1200 for X. So we just have to keep that in mind 1.1 bye bye to times and again, we have 1200 not 1600 1200 squared because excess squared on that term of the polynomial that we got from integrating. And no here again, just plugging in and plugging in 1200 and remembering that here the power is to the third. And then we have one more term, which is keep it in the same order is zero and then four fours for divided by for times 1600 out. I was made a mistake. 1200 because we're plugging in for the minus, which is the lower bound, which is 1200. And for this one, it's X to the fourth to the fourth and, uh, bring this out. And that equals what we wanted to find, which was dealt a sea going from 1200 to 1600 and plugging that into the calculator. You get that answer, which, All right down for you, which is 438 6693 All right, 0933 And then 0.3 repeated dollars. Is it asked for an answer in dollars on That's the problem. Gonna clean up the commas because I realized I did not do those correctly. And yeah, So the answer is 43 million, um, 866 1009 133.3. Repeating dollars is the difference in cost of production, um, going from producing 1200 units to 1600 units. So we applied the definition of an integral, which is saying that the integral of some function is the on bounds. A to B is the anti derivative of that function. Evaluated the bound, um at the upper bound, minus the anti derivative of that function at the lower bound. And then so we applied that to the formulas were given and for integrating. Um polynomial is which is what we had to do. We use the power rule which has stated here that just says the integral of X to the A and I just realized again, I've for gotten DX the integral of x t v a d X equals X to the A plus one divided by a plus one where a is constant. So applying those two ideas on doing a little bit of algebra and plugging into a calculator, you will get your answer of 43 million 100 66,933 1000.3 repeating dollars


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