Problem

(a) Show that if the profit $ P(x) $ is a maximum…

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Problem 59 Hard Difficulty

(a) If $ C(x) $ is the cost of producing $ x $ units of a commodity, then the average cost per unit is $ c(x) = C(x)/x $. Show that if the average cost is a minimum, then the marginal cost equals the average cost.
(b) If $ C(x) = 16,000 + 200x + 4x^{3/2} $, in dollars, find (i) the cost, average costs, and marginal costs at a production level of $ 1000 $ units; (ii) the production leve that will minimize the average cost; and (iii) the minimum average cost.

Answer

a)
And therefore, $C^{\prime}(x)=c(x)$
b)
(i) $342,491,342.491,389.74$
(ii) 400
(iii) 320

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Video Transcript

Okay here we are going to show that the marginal cost is equal to the average cost when average cost is a minimum. And then we are going to take a example where we are given a function for the cost of producing units and then find various attributes of the cost given some units. And then also minimize that function. So first we have to show that average cost and marginal costs are the same when average cost is at a minimum. So we're gonna do that using our um caution rule. So here's the formula for average cost is given in the problem. And so we want this to be at a minimum and to do that it has to be at a critical point. And so the derivative must be zero. So we're going to set let's change color here. We're in a set Derivative of average cost is equal to zero. And so that means derivative of cost which I marked with a big C. With like a little tip at the top to differentiate it from a normalcy. This is also equal to zero. So let's use our coercion rule. So derivative of the top times the bottom minus diverted of the bottom. That's just one. So I'm not gonna write anything times the top all over the bottom squared And this is equal to zero. So we're going to cancel the bottom real quick and simply add the cost to the other side and then divide by X. So if we do all that in order we get derivative of cost or the marginal cost because the marginal cost is the cost to add another unit of production. So it's like the growth rate of the cost and this is going to be equal to the cost over units produced, which of course is equal to our average cost. And so there's part A done using product role. Now we're going to tackle part B here. So first we have to find the cost, average cost and marginal cost when production level is 1000 units A K A when X is 1000 units. So cost is easy enough. We just have to plug in 10,000 here. And so if we let's do, I might need more room here, Let's do cost is going to be equal to 16,000 Plus 200 times 1000 is just 200,000 plus four times 1000 to the 3/2. So I'm going to write, this is 10 to the three, three halves, which does not cancel as well as I'd like. So let me just compute this out. Well, the calculator is able to show, but if you just compute this out, it's going to be 247 six, .77, repeating. So let's just round it up to 78. So this is how much it costs To produce 1000 units. Put a dollar. Sign out here, let me move this down. No way I can just move this down for now. And so now we have to compute the average cost. And so all we have to do to find average costs. If you remember this is cost, total cost. Let me use a different color Again, average costs is equal to total cost, which we just computed over X. So this is going to be 2476, We take out the comma 6 to 2.78 Divided by 1000, which will be equal to you. Just shift the decimal three points 247 And all around to 62 cents. And so finally we have to find the marginal cost. And if you remember in part a we showed that the average cost is equal to the marginal cost, but that's only when the average costs at a minimum. So we can't use that. So what we have to do to find marginal costs, let me drag this further down because I don't have enough room. We're going to take the derivative of our cost function. Remember the derivative of our cost function is the marginal cost. I don't know why I changed car. So 16,000 will go away because it's a constant 200 x. The root of 200 x is just 200 derivative of four X to the 3/2 is going to be six X to the 1/2 you take down the three halves, three have since four is six. And so all we have to do is plug in 1000. Again, this is our marginal cost derivative of cost is marginal cost. So 200 Plus six times the square root of 1000. And this will be equal to let me calculate is equal to three the dollar. Sign up 38, nine point seven will round up to four. So $389.74. And so that's part one. Now for part two and 3 of part B we have to find the production level or X. That will minimize average cost and then compute the actual average cost. So we have to take our function, Let me rewrite it down here. 16,000 Plus 200 x plus four X To the 3/2. There we are. And again be careful we have to minimize average cost. This is our normal costs or are total costs formula. So our average cost is going to be this divided by X. So we're going to have 16,000 divided by X plus 200 Plus four X to the 1/2. And so now to minimize this, we're just going to take the derivative. So derivative of average costs. Michael Sandel, this is going to be this will be negative 1600 16,000, sorry over X squared 200 goes away because it's a constant plus take down the one half, two X to the negative one half. Or we can write it as to over square root of X. So there we are. And now We want to set this equal to zero so to equal to zero and this will allow us to find the critical points. So we're gonna get to over square root X equals 1600. 16,000. Sorry. Over X squared. Multiply by X squared on both sides. We'll get two X squared is equal to sorry, two extra 3/2ves is equal to 16 16,000, divide by two extra three halves Equal to 8000. And so if you saw for X this is just equal to 400. Yes. So this is the value at which the average cost is minimized. So that's part two of part B. And so now for part three of part B we have to actually find the minimum average cost. So all we're gonna do is plugging this value for X. Whoa, plug in this value for X into our average cost formula. And then we will get the minimum average cost. So I'm going to call it C. Men This is equal to 16,000 Over against 400. 400 plus 200 Plus four times the square root of 400, Thankfully 400 is a nice square number. So when we simplify we get 400 plus 200. Hold on. Alright, this should be 40. So you get 40 plus 200 Plus four times square to 400 which is 24 times 20, is 80. And finally we arrive in our answer, which is 320. So this is the I guess we should put a dollar sign out here. This is the minimum average cost for this cost function.

Patrick D.

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Problem

(a) Show that if the profit $ P(x) $ is a maximum…

Problem 59 Hard Difficulty

Answer

a)
And therefore, $C^{\prime}(x)=c(x)$
b)
(i) $342,491,342.491,389.74$
(ii) 400
(iii) 320

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a)And therefore, $C^{\prime}(x)=c(x)$b)(i) $342,491,342.491,389.74$(ii) 400(iii) 320

Calculus: Early Transcendentals

Anna Marie V.

Heather Z.

Kayleah T.

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Integration Review - Intro

Definite vs Indefinite

Anna Marie V.

Heather Z.

Kayleah T.

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a)
And therefore, $C^{\prime}(x)=c(x)$
b)
(i) $342,491,342.491,389.74$
(ii) 400
(iii) 320