For Sammi Jo

Question: A company knows that unit cost C(x) and unit revenue R(x) from the production and sale of x units are related by:

    \[C(x)=\frac{R^2(x)}{120000}+11439\]

Find the rate of change of revenue per unit when the cost per unit is changing by $15 and the revenue is $2000.

Answer: So the above equation serves as the relation, this is the equation we’ll need to differentiate in order to obtain some related rates, which will allow us to solve the problem. We now know that:

    \[\frac{dC}{dx} = 15\]

and our goal is to solve for \frac{dR}{dx}. The question also states the revenue is $2000. First we must differentiate the equation relating C(x) and R(x):

    \begin{align*} \frac{C(x)}{dx} &= \frac{R^2(x)}{120000}+11439 \\ \frac{dC}{dx} &= \frac{d}{dx} \cdot \frac{1}{120000} \cdot R^2 + \frac{d}{dx} \cdot 11,439 \\ \frac{dC}{dx} &= \frac{1}{120000} \cdot \left( 2R \cdot \frac{dR}{dx} \right) + 0 \\ \frac{dC}{dx} &= \frac{dR}{dx}\cdot R \cdot \frac{1}{60000} \\ \frac{dC}{dx} &= \frac{dR}{dx} \cdot 2000 \cdot \frac{1}{60000} \\ \frac{dC}{dx} &= \frac{dR}{dx} \cdot \frac{1}{30} \\ 15 &= \frac{dR}{dx} \cdot \frac{1}{30} \\ \frac{dR}{dx} &= 450 \\ \end{align*}

The first line is simply the equation we were given from the question. The second line we obtain by beginning to differentiate the first line; I replaced the C(x) and R(x) notation with simply C and R to facilitate simplicity. \frac{1}{120000} is a scalar multiple so it can be pulled outside and is unaffected by differentiation. We then use the chain rule on R^2. Since 11439 is a constant, its derivative is simply 0. The next line is simplification. We then replace R with 2000 which was given in the question. Then we simplify further and solve for \frac{dR}{dx} and that’s that!

Alone With The Baby – Part II

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Alone With The Baby – Part I

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