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

1. 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 $\frac{dC(x)}{dx}_{x=50}=-30$

2. anonymous

thanks alot