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The cost (in dollars) of producing $ x $ units of a certain commodity is $ C(x) = 5000 + 10x + 0.05x^2 $.

(a) Find the average rate of change of $ C $ with respect to $ x $ when the production level is changed (i) from $ x = 100 $ to $ x = 105 $ (ii) from $ x = 100 $ to $ x = 101 $

(b) Find the instantaneous rate of change of $ C $ with respect to $ x $ when $ x = 100 $. ( This is called the \textit{marginal cost}. Its significance will be explained in Section 3.7.)

a) (i) 20.25$/$ unit, (ii) 20.05$/$ unitb) 20 $\mathrm{dollars} / \mathrm{unit}$

07:03

Daniel J.

01:50

Tyler M.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 7

Derivatives and Rates of Change

Limits

Derivatives

Campbell University

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Idaho State University

Boston College

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X. X. The amount of a commodity being produced Cfx is equal to 5000 plus 10 times X. Plus point oh five times X squared. We want to find the average rate of change in the cost function when ex uh goes from 100 to 105 and then we're gonna do it again when X. Goes from 100 to 101. Uh So what we have to do first is go ahead and find out what is the cost when X. Is 100 100 and 501. So we'll simply just put those in the calculator real quick substituting uh 100 in wherever you see X. So 5000 plus 10 times 100 plus 1000.5 times X. Squared. Um So I'll do that real quick and be right back. Sure. All right. So if we want the average rate of change in the function see Uh as as X. goes from 100 to 105 we have to take the change in sea and divided by the change in X. So the change in c. 6601 0.25 -60-500 Over to change an ex while X. is going from 100 to 105. That changed the next is five. Uh so the change in C. is going to be uh 66 01.25 -500. It's going to be 101.25 divided by five. And so 101 2025 divided by five gives us 20.25. So the average rate of change will be 20.25 uh dollars uh per unit of commodity. Alright, so to find the average rate of change in the cost function uh Between when x goes from 100 to 101 uh same kind of thing. Change and see over the change in x. Alright, when x goes from 100 to 101, The change in X is one. Uh the cost uh change in the cost when X is 100 compared to 101 is the difference between 60 500 And 65, 20.05. So basically just $20.05. And so the average rate of change when x goes from 100 to 101 is going to be $20.5 per unit. Lastly we want to find the instantaneous rate of change. Uh When X is 100 instantaneous rate of change is the derivative. So we have to find, see prime of X. Take the derivative of the cost function. While the derivative of 5000, that's a constant. That zero, the derivative of 10 x is simply 10 plus. Then the derivative of point oh five times X squared while the derivative of X squared is two X. So times point oh five Is going to be .10 x. So that is your derivative. See prime of X. And we if we want to find the value of that derivative when X is 100 we have to evaluate it. The next is 100. We simply just plug 100 for X. So see prime of X is 10 plus 0.10 times X. But we're substituting 100. And for exa 1000.10 times 100 Well .10 times 100 is 10, so 10 plus 10 is 20. So the cost function is changing at a rate of $20 per unit. When X is 100 that is instantaneous rate of change. Notice the average rate of change at 100 when we calculated 100 to 101 was $20.05. When we calculated the average rate of change from 100 to 105 we got $20.25 per unit. But the instantaneous rate of change at 100 is exactly $20 per unit.

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