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Problem 31 Medium Difficulty

Suppose that the cost (in dollars) for a company to produce $ x $ pairs of a new line of jeans is
$ C(x) = 2000 + 3x + 0.01x^2 + 0.0002x^3 $
(a) Find the marginal cost function.
(b) Find $ C'(100) $ and explain its meaning. What does it predict?
(c) Compare $ C'(100) $ with the cost of manufacturing the 101st pair of jeans.

Answer

a) $3+0.02 x+0.0006 x^{2}$
b) C'(100) = $11$ is the rate at which costs are increasing with respect to the production level when $x = 100. C'(100)$ predicts the cost of producing the 101st pair of jeans.$C'(x) = 3 + 0.02x + 0.0006x^2$
c) $C^{\prime}(100)=C(101)-C(100)$

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Video Transcript

All right, So we have a business application here, and the first thing we're asked to do is find the marginal cost and the marginal cost is the derivative of the cost function. So we take the derivative of the given function, and we get three plus 0.2 x plus 0.6 X squared for part B. We want to find the marginal cost for 100. So we go ahead and weigh substitute 100 into this equation. The numbers work out pretty nicely. You could even do it without a calculator if you really wanted to. And you end up with 11. So what that means is, if you've already made 100 items, that would be the cost to make the 101st item. That's an estimate of it. At least, using the derivative gives you an estimate, and the item in this case would be a pair of jeans, okay. And for part c, what we're asked to do is find the actual cost of manufacturing the 101st pair of jeans and compare that to what we found for part B, The estimate. So to get the at actual cost of that 101st pair, we're gonna have to take the cost of 101 pairs and subtract the cost of 100 pairs. And that way we get the cost of that one additional pair. And what we're going to do is use a calculator to find each of these costs. And for 101 pairs, we get $2611.7. And for 100 pairs of jeans, we get $2600. So when we subtract those we get $11.7 was the actual cost of producing that 101st pair of jeans that is very close to the estimated cost.