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anonymous
 4 years ago
Suppose a manufacturer finds that its average weekly costs of production, C (in thousands of dollars), are a function of the number of units produced, x (in hundreds).
If C(x) = 0.004 x^30.6 x^2+500,
then
dC/dx x=50
anonymous
 4 years ago
Suppose a manufacturer finds that its average weekly costs of production, C (in thousands of dollars), are a function of the number of units produced, x (in hundreds). If C(x) = 0.004 x^30.6 x^2+500, then dC/dx x=50

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ash2326
 4 years ago
Best ResponseYou've already chosen the best response.1we have \[C(x)=0.004x^30.6x^2+500\] take the derivative with respect to x \[\frac{dC(x)}{dx}=0.012x^21.2x+0\] let's substitute x=50 to find fdC(x)/dx at 50 \[\frac{dC(x)}{dx}_{x=50}=0.012(50)^21.2*50\] \[\frac{dC(x)}{dx}_{x=50}=0.012*25001.2*50\] we get \[\frac{dC(x)}{dx}_{x=50}=30\]
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