Does anyone know the answer (s) to this?

[Jimmy Perdon]

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]

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.[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-10,"ymin":-7.65696784073507,"xmax":10,"ymax":7.65696784073507}},"randomSeed":"86f5ca9aa79de8cb0e8d39ee14441c5a","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"P=2Av^{2}+\\frac{B}{v}"},{"type":"expression","id":"3","color":"#388c46","latex":"B=-1","hidden":true},{"type":"expression","id":"2","color":"#2d70b3","latex":"A=-1","hidden":true}]}}[/DESMOS]

 

[skeeter]

https://mathhelpboards.com/threads/does-anyone-know-the-answer-to-this.28654/