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(e) What is the minimum average cost?...

Question

(e) What is the minimum average cost?

(e) What is the minimum average cost?



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Minimum Average cost The cost of producing $x$ units of a product is modeled by $C=500+300 x-300 \ln x, \quad x \geq 1$ (a) Find the average cost function $\bar{C}$. (b) Analytically find the minimum average cost. Use a graphing utility to confirm your result.

Alright for this problem. We need these two key ideas here. The first is that the average just like whenever you guys think of average, the average cost is going to be the total cost divided by the number of units. Okay, so it's just like any average you take average test score, get your total test score divide by the number of tests here. We're going to need our total cost divided by how many units we made. Okay. And then the second key is one of the big ideas in calculus. If you want to minimize something, what you're going to need to figure out is when is the derivative equal to zero? And when is the derivative undefined? Okay. So this problem asks us to find the minimal average cost when we know the total cost. So it's a little cumbersome. But the first thing we need to do is find the average cost and then we need to find the minimum. So we need to find when the derivative of that is zero. Okay. It's a little cumbersome but actually it's not not too bad at all. Okay, so the first part part A is we want to find the average cost which we're going to denote by C. Bar. Okay. And so again that's just going to be our total cost. Which is C. In this problem divided by the number of units which is X. Okay. And then from the problem we know the total cost of X units is given by this equation. So that is going to be the c. That's a 100 Plus the 25 x -120 Natural Log of X. And then I'm going to divide that by X. Okay. And again, that is our average cost. Super. Not too bad at all. Total divided by the number that we have. Okay, great. Now, if I want to minimize that, I need to do this, I need to find its derivative. Figure out when it's zero and when it's derivative is undefined. So to do that, I'm going to simplify this average cost here a little bit. Okay, I can rewrite this as 100 over X Plus 25 x over X. Okay, I'm doing this basically because I don't want to use the quotient rule if I don't have to or if I do need to use the question wrong, I'm going to use it on something small. Okay, okay. And now this simplifies a bit this first term here, that is going to be the same as mhm. Just 100. And then instead of dividing by ex let me rewrite that as X to the -1. The second term here, those exes cross out And I'm going to get plus 25. And then this third term here minus hopes I forgot an X. Here. That was an ex Okay. 120 times the natural log of X divided by X. Well that I'm going to just leave as a quotient, but that's not too bad. It's a pretty simple one. Okay. Alrighty. And again, I just did that to make finding my derivative a little easier. And the reason I need to find the derivative is again, they want me to find the minimal average cost. And so again, like it says up here to minimize something, you need to look at the derivative. So I need to calculate C. Prime first, find when at zero and find when it's undefined. Okay, so let's take the derivative Derivative is gonna be negative 100 X to the negative. Too Derivative of 25 0 derivative of that last part here, we're going to need a quotient rule to find the derivative of that. So that is the bottom the X derivative of the top is 1 20 times the derivative of natural log is one over X. There's my bottom derivative of the top minus. The top 1 20. Natural log of X Times the derivative of the denominator, which is one all over the denominator squared. Okay. Alrighty. So let's simplify that a little bit here. We have negative 100 over X squared Plus zero. So that's not going to change anything minus. And now X times 120 times one over X is just going to be 1 20 -120 Natural Log of X. All that over X squared. Okay. And so I can simplify that a little bit. Let me do negative 100 minus. And now careful algebra thing coming up here, This negative side goes to that 120 and that negative 120. So this is going to be -120 plus 1 20 X or 1 20 natural log of X. And what I'm doing here is I have fractions with a common denominator, so I'm just going to add those fractions. Okay, super. So again have X squared and X squared in the denominator. And so I'm just going to add those two fractions together. Okay, and now when I do that I can subtract the 100 the 1 20 get negative 2 20 plus 1 20 natural log of X. All divided by X squared. Okay. Yeah, awesome. So that's my derivative. Now let me square myself off a little bit here. Okay. To minimize the questions I ask are again one is the derivative equal to zero and when it's a derivative undefined, So the derivative is zero when the numerator is zero. So that's when negative. 220 plus the negative natural log of X. Excuse me? When is that zero? And that will make my derivative equal to zero. And then when is my denominator equal to zero. Okay. All right. So To solve this first one, when the derivative is zero, going to add to 20 to both sides and I forgot it again. Awesome. So let me erase because this isn't just a natural log of X here. If we check over here, I always forget this 120 here. So plus 120 natural log effects. Okay, So I'm still going to add to 20 to both sides. I'm going to be left with 120 natural log of X. Again that 100 twenty's coming from right here Equals 220, Divide both sides by 120. So the natural log of X Is to 20/120. Okay. And now to solve that, what we're gonna do is we're going to raise both of E to both of those powers. So E to the natural log of X Is E to the 220 Over 120. And when I do E two, the natural log of something that just gets me an X. Ok. Those are my new properties of E exponentials and logarithms. And over here, this is a calculator problem. Right? If I do eat to the to 20/1 20 make sure you do that. And this whole thing is in the Exponent. This whole guy is in the exponents. You might need to use parentheses to do that. You should end up with 6.2 55 Okay. All right. So that's when my derivative is zero. When is my derivative undefined? Well, over here, that's when X squared is zero. So that would tell Me X0. Okay. But the problem states that X is always greater than one because I want to make more than one unit. So the problem says X is greater than one. Okay, So X is Never zero. So I don't get any points from this. Okay. So I really don't have to think about this at all. Okay. Alrighty. Again, the only time I would get something from this is if the X is zero but my ex has to be bigger than one. So I don't get anything. Okay. Okay. Now the last little fun part about this problem is up here, what it asks us for, it asks us to find the minimal average cost. And what I have so far down here is how many units I need to make 6.255. So to find the minimal average cost, I need to plug that in two. The um, minimal average cost, which is right, Sorry. It wants me to find yes. The average cost, the minimal average cost, which is this guy right here. Okay. So that is what I need to plug into. The thing I took the derivative of and X is the number of units it wants the cost. Okay. So I need to take and find see Bar evaluated when X is 6.255. Okay, So I'm going to do a 100 plus 25 Times 6.255 minus 1, 20 Natural Log of 6.255. Do all of that. And then divide the answer by X, which is 6.255. Alternately, if you guys wanted to, you could have plugged it into this simpler version. Doesn't should get the same answer. It doesn't matter which one you do. Okay. And when you plug this in, you should get an answer of 5.81. Okay. And again, that's going to be it's an average, so it's going to be dollars per unit. Okay. Great problem. Have to find the equation. And then remember, what does it mean to minimize something? We have to take the derivative and set it equal to zero. Great. Thank you.

In the long run we're talking about when all costs are variable and the long run average cost curve basically denotes the points at which the firm can produce any given level output for the lowest cost. So the short run average cost curves can show us how much we can produce at any given coin. Ah, point. So this will be quantity. This is price at any given point, any given quantity. This is the lowest price at which we can produce anything. And the longer on average cost curve basically just shows you at which of those points you can or is the aggregation of all the minimum point. So this is really just 1,000,000 or an infant number of short run average cost curves tangential to the longer an average cost carriers.

So in this question, we want to fight a production level that minimizes the average cost where excess measured in thousands of units. So we do this by setting emcee equal to a C. We find M. C as a derivative which is M C equals the drift of this X eat of the explosive excuse in the product rule minus four X. We set this equal to the average cost, which is just see if x divided by ex So either the X minus two x this equaling average cost. So we want to get rid of us. Eat of the ex cons some both sides. Now we can add two X It'll side to him. The X eat of the X equals two X. You get either the X equals two and we can get expert taking that Ellen Ellen of two. So to test that this is really the minimum, we can use the second derivative test on average cost, which is just CIA banks over X, which we found right here. If we take the German again, this is the derivative of this. If we take the derivative twice, we get either the ex and then either The ex twice forget this. So either the ex is going to be positive for all of for Ellen of two. And so that means that this really is the minimum. And of course, X is measured in thousands of units. So we do 1000 times. Ellen of two thiss gives us roughly 693 693 units, the production level that minimizes average cost.

The number forty six and six point one. I'm going to pull up the question so we can look at it, find the minimum value so we know we're minimizing this function of the average cost. So let's stop right there. Average cross that we want to. Well, we want to minimize the average cost. Average cost is our sea bar. See bar is equal to the cost function divided by X For us, that's equal to seventy five x squared plus twenty x plus thirty six seventy five divided by X. We're going to rewrite this as seventy five x plus twenty plus thirty six seventy five times, one over X. Now we can go back to the question. We know we're trying to minimize that cost function, and we want to do it over the following intervals. So we have a is one to ten. B is from ten to twenty. I left some notes you'LL notice over here the first thing we need to do to find the minimum finding test the critical points which must be in those given bounds and also test the in point. So here's the function we want to minimize our first step is going to be to find the critical points, which means we need to find the derivative. So this is like part one. We need to find C bar, prime of X. Okay. And see bar Prime of X is going to be seventy five minus thirty six seventy five over X quips takes some time. Make sure you can find that to rivet upon your own to find critical points we invest set this equal to zero. That means we have that negative thirty six seventy five over X squared equals negative seventy five. Which means if we do some cross multiplication, you're going to end up with X squared is equal to thirty six seventy five, divided by seventy five X squared is equal to forty nine X is equal to plus or minus seven. These are the critical points and we're going to use those for both a and be in this question. So now we're going to look at part eh? Per is one less than or equal to x lesson or equal to ten. And if you recall the steps I wrote at the first page, we always need to test the bounds. One in ten in any critical numbers, but the critical numbers must be in this range. Negative seven is not in the range. We're not going to test that. We're plugging this into our average cost function. The's excise playing in one average crosses three thousand seven hundred and seventy playing in ten, one thousand, one hundred and thirty seven point five and seven is one thousand and seventy. Ok, that looks pretty clear here that the minimum average cost is ten. Seventy and that is at X equals seven. Kirby is just changing our interval ten was center equals X lesson or equal to twenty. If we go to the beginning of the steps I wrote, we always test are in point ten and twenty and we test our critical points. However negative seven and seven, neither of them are in these bounds. So we're not even gonna worry about the critical point we're on ly going to plug in our in points to the average cost. Okay. Putting in ten, one thousand one hundred and thirty seven point five zero when x is twenty and you plug it in tow. Average cost you She at one thousand seven hundred and three point seventy five. So now we have that the minimum average cost is one thousand one hundred and thirty seven dollars and fifty cents, and that is that X equals ten. So this is a really good example of showing you how important your bounds are, Okay? Because you can on ly test critical points that fall in your domain or range or your bounds here on X.


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