Let's carefully address part (b) of the question using the information from the image:

Part (b): Prove using $\frac{d^2C}{dv^2}$ that the cost is minimized at the speed found in (a)(i).

From part (a)(i), the cost function is: $C = \frac{1500}{v} + \frac{2v}{11} + 60$

We already know: $\frac{dC}{dv} = -\frac{1500}{v^2} + \frac{2}{11}$

At the minimum cost, $\frac{dC}{dv} = 0$, which gave: $\frac{1500}{v^2} = \frac{2}{11}$ Solving this, we found ( v = \sqrt{\frac{1500 \cdot 11}{2}} = \sqrt{8250} \approx 90.83 , \text{km/h.} )

Now, calculate $\frac{d^2C}{dv^2}$:

The second derivative of $C$ is: $\frac{d^2C}{dv^2} = \frac{d}{dv}\left(-\frac{1500}{v^2} + \frac{2}{11}\right)$

Differentiate term by term: $\frac{d^2C}{dv^2} = \frac{d}{dv}\left(-\frac{1500}{v^2}\right) + \frac{d}{dv}\left(\frac{2}{11}\right)$

The derivative of $-\frac{1500}{v^2}$ is: $\frac{d}{dv}\left(-\frac{1500}{v^2}\right) = 3000v^{-3} = \frac{3000}{v^3}$

The derivative of $\frac{2}{11}$ is: $\frac{d}{dv}\left(\frac{2}{11}\right) = 0 \, \text{(since it's a constant)}.$

Thus: $\frac{d^2C}{dv^2} = \frac{3000}{v^3}.$

Evaluate $\frac{d^2C}{dv^2}$ at $v = 90.83$:

Substitute $v = 90.83$ into $\frac{3000}{v^3}$: $\frac{d^2C}{dv^2} = \frac{3000}{(90.83)^3}.$

First, calculate $(90.83)^3$: $(90.83)^3 \approx 748024.49.$

Now calculate: $\frac{3000}{748024.49} \approx 0.00401.$

Conclusion:

Since $\frac{d^2C}{dv^2} > 0$ (positive value), this confirms that the cost function $C$ has a minimum at $v = 90.83 \, \text{km/h.}$

Do you have any questions or need further clarification? Let me know!

5 Related Questions:

1. What is the significance of the second derivative test in optimization problems?
2. How does the speed $v$ affect the cost function $C$ in this model?
3. Can this method be applied to non-polynomial cost functions? Why or why not?
4. What are the physical interpretations of $\frac{dC}{dv}$ and $\frac{d^2C}{dv^2}$ in this context?
5. How would the results change if an additional term was added to the cost function?

Tip:

Always verify the units of each term in a cost function to ensure they are consistent, especially in applied mathematics problems like this!