Electric Delivery Van Highway Cruising Optimization (Mechanical Engineering) Solution

Problem Overview

A mechanical fleet engineer optimizes the cruising speed for electric delivery vans traveling a $240\text{-mile}$ round trip. Labor costs $\$45/\text{hr}$ and electrical battery consumption costs $\frac{v^2}{80}\text{ \$/hr}$ on the legal domain $v \in [45, 75]\text{ mph}$. We formulate the single-variable total cost function, find the optimal cruising velocity using calculus, verify the global minimum via the Closed Interval Method, and validate the Cost Equipartition Theorem.

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Step-by-Step Mathematical Solution

Step 1: Formulate Transit Time and Total Cost Objective Functions

The modeling process follows the Applied Optimization Modeling Framework: (Defined in Applied Optimization Modeling Framework.)

1. Total Transit Time: For a constant cruising speed $v$ (in $\text{mph}$) over round-trip distance $D = 240\text{ miles}$:

   $$t(v) = \frac{D}{v} = \frac{240}{v}\text{ hours}$$

2. Total Trip Cost Objective Function: Total operating cost is the product of transit time and the combined hourly burn rate:

   $$C(v) = t(v) \cdot \left[ R_{\text{labor}} + R_{\text{energy}}(v) \right] = \frac{240}{v} \left( 45 + \frac{v^2}{80} \right)$$
   Distribute the $\frac{240}{v}$ term:
   $$C(v) = \frac{240 \times 45}{v} + \frac{240 v^2}{80 v} = \frac{10,800}{v} + 3v\text{ dollars}$$
   Rewrite in negative exponent form for direct power-rule differentiation:
   $$C(v) = 10,800 v^{-1} + 3v$$

Step 2: Establish the Context-Driven Domain

Fleet speed governor restrictions and Interstate-15 legal speed limits establish the physical domain:

Because $C(v)$ is continuous on the closed, bounded interval $[45, 75]$, the Extreme Value Theorem guarantees the existence of an absolute minimum.

Step 3: Differentiate and Find Critical Numbers

Differentiate $C(v)$ with respect to velocity $v$:

Set $C'(v) = 0$ to identify stationary points:

Taking the positive physical velocity: Because $60 \in [45, 75]$, $v = 60\text{ mph}$ is a valid critical point.

Step 4: Execute the Closed Interval Method

Evaluate $C(v) = \frac{10,800}{v} + 3v$ at the boundaries and the interior critical point:

1. At Minimum Governor Speed ($v = 45\text{ mph}$):
   • Trip duration: $t = \frac{240}{45} \approx 5.333\text{ hours}$.
     $$C(45) = \frac{10,800}{45} + 3(45) = 240 + 135 = \$375.00$$
2. At Maximum Speed Limit ($v = 75\text{ mph}$):
   • Trip duration: $t = \frac{240}{75} = 3.200\text{ hours}$.
     $$C(75) = \frac{10,800}{75} + 3(75) = 144 + 225 = \$369.00$$
3. At Critical Cruising Speed ($v = 60\text{ mph}$):
   • Trip duration: $t = \frac{240}{60} = 4.000\text{ hours}$.
     $$C(60) = \frac{10,800}{60} + 3(60) = 180 + 180 = \$360.00$$

Comparing the three candidate evaluations:

The absolute minimum operational trip cost is $\$360.00$, occurring at a cruising velocity of $60\text{ mph}$.

Step 5: Verification of the Cost Equipartition Theorem

At the optimal speed of $v = 60\text{ mph}$:

• Total Labor Cost: $C_{\text{labor}} = \frac{10,800}{60} = \$180.00$
• Total Battery Charging Cost: $C_{\text{energy}} = 3(60) = \$180.00$

The labor cost and energy charging cost are exactly equal ($C_{\text{labor}} = C_{\text{energy}} = \$180.00$), mathematically confirming the Cost Equipartition Theorem formalized in Cruising Velocity and Operational Transit Cost Optimization.

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Engineering Significance & Conclusion

• Operational Recommendation: Fleet dispatchers should configure commercial cruise control governors to $60\text{ mph}$ on this route. Operating at $60\text{ mph}$ saves $\$15$ per truck shift compared to driving at $45\text{ mph}$, and saves $\$9$ per shift compared to driving at $75\text{ mph}$. For a 100-vehicle fleet completing 300 shifts annually, this optimization generates $\$270,000$ in annual operational savings.
• Impact of Wage Inflation: If driver wages increase (e.g., from $\$45/\text{hr}$ to $\$60/\text{hr}$), the numerator of the time-cost component increases ($wD = 60 \times 240 = 14,400$), shifting the optimal speed:
  $$v^* = \sqrt{\frac{14,400}{3}} = \sqrt{4,800} \approx 69.3\text{ mph}$$
  As labor becomes more expensive, the company saves money by driving faster to reduce total paid hours, even though aerodynamic battery drag increases.