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Mathematics 39 Online
OpenStudy (anonymous):

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^3-0.6 x^2+500, then dC/dx x=50

OpenStudy (ash2326):

we have \[C(x)=0.004x^3-0.6x^2+500\] take the derivative with respect to x \[\frac{dC(x)}{dx}=0.012x^2-1.2x+0\] let's substitute x=50 to find fdC(x)/dx at 50 \[\frac{dC(x)}{dx}_{x=50}=0.012(50)^2-1.2*50\] \[\frac{dC(x)}{dx}_{x=50}=0.012*2500-1.2*50\] we get \[\frac{dC(x)}{dx}_{x=50}=-30\]

OpenStudy (anonymous):

thanks alot

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