MHB Mylesbibbs' question at Yahoo Answers regarding marginal cost

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The cost function for producing automobile tires is given as C(q) = 2300 + 15q - 0.01q², where q represents the number of tires produced. To find the rate of change of cost with respect to time, the chain rule is applied, resulting in dC/dt = (15 - 0.02q)(dq/dt). With dq/dt set at 30 tires per week and q at 400, the calculation shows that dC/dt equals $210. Therefore, the production cost is increasing at a rate of $210 per week.
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Here is the question:

CALC. HELP! Related Problems!?!?

The weekly cost C, in dollars, for a manufacturer to produce q automobile tires is given by
C = 2300 + 15q − 0.01q2 0 ≤ q ≤ 800.
If 400 tires are currently being made per week but production levels are increasing at a rate of 30 tires/week, compute the rate of change of cost with respect to time.

Here is a link to the question:

CALC. HELP! Related Problems!?!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello mylesbibbs,

We are given:

[math]C(q)=2300+15q−0.01q^2[/math]

and we are asked to find [math]\frac{dC}{dt}[/math]

Now, if we take the given cost function and differentiate with respect to time, using the chain rule we find:

[math]\frac{dC}{dt}=\frac{dC}{dq}\cdot\frac{dq}{dt}= \left(15-0.02q \right)\frac{dq}{dt}[/math]

We are also told that [math]\frac{dq}{dt}=30[/math] and so we have:

[math]\frac{dC}{dt}=30(15-0.02q)[/math]

and so at the current production level of $q=400$, we find:

[math]\left.\frac{dC}{dt}\right|_{q=400}=30(15-0.02\cdot400)=210[/math]

Thus, we find that the production cost is increasing at a rate of \$210 per week.

To mylesbibbs and any other guests viewing this topic, I invite and encourage you to post other marginal cost questions here in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.
 
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