We are tasked with minimizing the cost $C$ given by the equation:
$C = \frac{720}{x} + 0.2x + 6,$ where $x$ represents the speed of the car in mph.

Step 1: Differentiate the cost function

To find the value of $x$ that minimizes $C$, we compute the first derivative of $C$ with respect to $x$: $C' = \frac{d}{dx} \left( \frac{720}{x} + 0.2x + 6 \right).$ Using the power rule and the derivative of $\frac{1}{x}$, we get: $C' = -\frac{720}{x^2} + 0.2.$

Step 2: Solve for critical points

Set $C' = 0$ to find the critical points: $-\frac{720}{x^2} + 0.2 = 0.$ Rearrange to isolate $\frac{720}{x^2}$: $\frac{720}{x^2} = 0.2.$ Multiply through by $x^2$ and divide by $0.2$: $x^2 = \frac{720}{0.2} = 3600.$ Take the square root of both sides: $x = \sqrt{3600} = 60.$

Step 3: Verify minimum with the second derivative

Compute the second derivative of $C$: $C'' = \frac{d}{dx} \left( -\frac{720}{x^2} + 0.2 \right) = \frac{1440}{x^3}.$ Substitute $x = 60$ into $C''$: $C'' = \frac{1440}{60^3} > 0.$ Since $C'' > 0$, the cost function has a local minimum at $x = 60$.

Step 4: Find the minimum cost

Substitute $x = 60$ into the original cost function: $C = \frac{720}{60} + 0.2(60) + 6.$ Simplify: $C = 12 + 12 + 6 = 30.$

Final Answer:

• The speed $x$ that minimizes the cost is $\mathbf{60 \, \text{mph}}$.
• The minimum cost $C$ is $\mathbf{30}$.

Let me know if you would like a more detailed explanation or further clarifications!

Related Questions:

1. How does the second derivative test confirm a minimum value?
2. Can you explain why setting $C' = 0$ gives critical points?
3. What happens to the cost as $x \to \infty$ or $x \to 0$?
4. Could this optimization problem change if additional terms were added to $C(x)$?
5. What is the significance of minimizing cost in real-world scenarios like this?

Tip:

Always check the second derivative to confirm whether a critical point is a minimum or maximum.