# Does anyone know the answer (s) to this?

#### Jimmy Perdon

##### New member
A simplified model of the power P required to sustain the motion of an electric car at speed v experiencing nonzero drag can be modeled by the equation:

P = Av^2 + (B/v)
(where A and B are positive constants.)

(a) What speed vP minimizes power?
(b) What power does the speed in (a) require?
(c) Suppose that an electric car has a usable store E of energy. How far dP can the electric car travel at the
speed found in (b)?

#### Greg

##### Perseverance
Staff member
Here's the derivative for the function of power, $$\displaystyle P=Av^2+\frac{B}{v} (\text{Where }A\text{ and }B\text{ are positive constants)}$$ that they give in the introduction to the problem:

$$\displaystyle P'(v)=2Av-\frac{B}{v^2}$$

To answer $$\displaystyle \text{(a) What speed }vP\text{ minimizes power?}$$ we set this expression equal to $$\displaystyle 0$$ and solve for $$\displaystyle v$$:

\displaystyle \begin{align*} 2Av-\frac{B}{v^2}&=0 \\ 2Av&=\frac{B}{v^2} \\ 2Av\cdot\frac{v^2}{B}&=\frac{B}{v^2}\cdot\frac{v^2}{B} \\ \frac{2Av\cdot v^2}{B}&=1 \\ \frac{2Av^3}{B}\cdot\frac{B}{2A}&=1\cdot\frac{B}{2A} \\ v^3&=\frac{B}{2A} \\ v=\sqrt[3]{\frac{B}{2A}} \\ \end{align*}

Now we need to determine if this result is a minimum or a maximum and we do this by examining the concavity of the graph of $$\displaystyle P$$

I'll let you take a look at the graph and experiment with the different values for $$\displaystyle A$$ and $$\displaystyle B$$. As a suggestion, take a look at the graph when $$\displaystyle A$$ and $$\displaystyle B$$ are different signs.

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