(a) If C(x) is the cost of producing x units of a commodity, then the average cos

(a) If C(x) is the cost of producing x units of a...

Key Concepts

Optimization and Critical Points

Optimization in this context involves finding the production level that minimizes the average cost. This is done by taking the derivative of the average cost function, setting it equal to zero, and solving for x, which identifies the critical points. These points indicate potential minima or maxima of the function.

Marginal Cost

Marginal cost is the derivative of the cost function, C'(x), and it represents the additional cost of producing one more unit. It is an essential concept for making production decisions and analyzing cost behavior at different production levels.

Relationship Between Marginal and Average Cost at the Minimum

By applying differentiation to the average cost function and setting its derivative equal to zero, one can show that at the point where the average cost is minimized, the marginal cost is equal to the average cost. This relationship is a key result in production analysis and cost optimization.

Cost Function

The cost function represents the total cost incurred in producing x units of a commodity. It typically includes fixed costs, which do not change with production volume, and variable costs, which depend on the number of units produced.

Average Cost

The average cost is defined as the total cost divided by the number of units produced (C(x)/x). It represents the cost per single unit and is crucial for understanding production efficiency and pricing strategies.

Transcript

00:01 For this problem, we're going to be looking at a few different kinds of cost for producing a commodity.

00:07 In particular, there's three that we're going to look at.

00:10 Let's start with just plain old cost.

00:12 The cost of producing x units of a commodity is capital c of x.

00:19 And i'm going to try to make my capital and lowercase very distinctive.

00:22 So capital c of x is our cost.

00:24 We are also told in the book, if you read through the textbook, something called the marginal cost.

00:31 And the marginal cost can be found by taking the derivative of c of x.

00:37 Our third thing that we're going to talk about for this problem is something called the average cost.

00:42 And we're told that the average cost is a lowercase c of x.

00:47 And that's going to equal the cost function, capital c of x, divided by x.

00:53 And for this first piece of this problem, we want to see what happens if i keep the average cost to a minimum.

01:00 I want to minimize my average cost c of x.

01:06 Well, anytime we're looking to minimize or maximize a function, we can do that by taking the derivative and setting it equal to zero.

01:14 That will give us our critical points where our maximum and minimums will happen.

01:19 So let's minimize c of x.

01:22 Now again, that's capital c of x over x.

01:25 If i want to take the derivative.

01:28 This is a quotient, so i need to use the quotient rule, which is the denominator times the derivative of the numerator, that's capital c of x, minus the numerator, capital c of x, times the derivative of the denominator.

01:44 In this case, that's just one, all over the denominator squared.

01:50 Now, again, minimizing a function, i'm going to take this and set it equal to zero.

01:55 Now, my denominator won't be zero because i'm producing some units.

02:01 I want x, i want to actually be producing something.

02:03 So x is not going to be zero, so the denominator is not zero.

02:07 So i want to set the numerator equal to zero.

02:11 If i do that, that's going to give me x times the derivative of my cost function, and i'm going to add the cost function to both sides, that capital c of x.

02:22 So this is what i have, x times the derivative of the derivative of the cost function.

02:26 My cost function equals the cost function.

02:30 Then for this problem, i want to see, i want to set this, i want to solve for the marginal cost, which is here with this green star derivative of the cost function.

02:42 So i am going to solve for capital c prime of x, and that equals capital c of x divided by x.

02:51 I have to divide both sides by x.

02:53 Well, if you notice, this is the red star, that is the average cost.

03:00 So when i keep the average cost to a minimum, the average cost equals the marginal cost.

03:10 So that's just something to keep in mind.

03:12 That'll help us as we're solving a problem in a little bit.

03:15 Okay.

03:16 Now that we've talked about those, let's look at the other pieces.

03:20 Let's do a little more intentional here.

03:23 I have an actual function instead of just doing generalities.

03:26 I have a cost function, capital c of x.

03:29 Equals 16 ,000 plus 200x plus 4x to the three halves.

03:42 So that's my cost function.

03:45 I would like to find the cost, the average cost, and the marginal cost when x is 1 ,000 units.

03:52 So here is cost.

03:55 So if i want to find the cost for 1 ,000 units, i'm just going to plug 1 ,000 in everywhere there's an x here.

04:02 If i do that and put that in my calculator, i'm going to end up with a cost of 342 ,491, 11 cents.

04:16 So that is my cost function evaluated when x is 1 ,000.

04:21 How about marginal cost? well, marginal cost, if you remember from our definitions, marginal cost is going to be the derivative.

04:30 So let's find the derivative of our cost function.

04:33 Well, derivative of 16 ,000 is just zero.

04:36 It's just a constant.

04:38 Then i have derivative of 200x is 200.

04:42 And 4x to the three halves.

04:44 I bring down that three halves.

04:46 4 times three halves is going to be 6x, and i subtract one that puts it to one half...

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