How Fast Should the Airplane Fly? The cost of operating a jet-powered commercial (passenger-carrying) airplane varies as the three-halves (3/2) power of its velocity; specifically, CO = knv^3/2, where n is the trip length in miles, k is a constant of proportionality, and v is velocity in miles per

Question

How Fast Should the Airplane Fly?
The cost of operating a jet-powered commercial (passenger-carrying) airplane varies as the three-halves (3/2) power of its velocity; specifically, [latex]C_{O}=knv^{3/2}[/latex], where n is the trip length in miles, k is a constant of proportionality, and v is velocity in miles per hour. It is known that at 400 miles per hour, the average cost of operation is $300 per mile. The company that owns the aircraft wants to minimize the cost of operation, but that cost must be balanced against the cost of the passengers’ time (C_{C}, which has been set at $300,000 per hour.

(a) At what velocity should the trip be planned to minimize the total cost, which is the sum of the cost of operating the airplane and the cost of passengers’ time?

(b) How do you know that your answer for the problem in Part (a) minimizes the total cost?

Accepted Answer

(a) The equation for total cost ([latex]C_{T}[/latex]) is

[latex]C_{T}=C_{O}+C_{c}=k v w^{3 / 2}+(\$ 300,000 \text { per hour })\left(\frac{n}{\partial}\right)[/latex]

where n/v has time (hours) as its unit.
Now we solve for the value of k:

[latex]\frac{C_{O}}{n} =k v^{3 / 2} \[/latex]

[latex]\frac{\$ 300}{\text { mile }} =k\left(400 \frac{\text { miles }}{\text { hour }}\right)^{3 / 2} \[/latex]

[latex]k =\frac{\$ 300 / \text { mile }}{\left(400 \frac{\text { miles }}{\text { hour }}\right)^{3 / 2}} \[/latex]

[latex]k =\frac{\$ 300 / \text { mile }}{8000\left(\frac{\text { miles }^{3 / 2}}{\text { hour }^{3 / 2}}\right)} \[/latex]

[latex]k =\$ 0.0375 \frac{\text { hours }^{3 / 2}}{\text { miles }^{5 / 2}}[/latex]

Thus,

[latex]C_{T}=\left(\$ 0.0375 \frac{\text { hours }^{3 / 2}}{\text { miles }^{5 / 2}}\right)(n \text { miles })\left(v \frac{\text { miles }}{\text { hour }}\right)^{3 / 2}+\left(\frac{\$ 300,000}{\text { hour }}\right)\left(\frac{n \text { miles }}{v \frac{\text { miles }}{\text { hour }}}\right) \[/latex]

[latex]C_{T}=\$ 0.0375 n \tau^{3 / 2}+\$ 300,000\left(\frac{n}{v}\right)[/latex]

Next, the first derivative is taken:

[latex]\frac{d C_{T}}{d v}=\frac{3}{2}(\$ 0.0375) n v^{1 / 2}-\frac{\$ 300,000 n}{v^{2}}=0.[/latex]

So,

[latex]0.05625 v^{1 / 2}-\frac{300,000}{v^{2}}=0. \ 0.05625 v^{5 / 2}-300,000=0 \ v^{5 / 2}=\frac{300,000}{0.05625}=5,333,333 \ v^{*}=(5,333,333)^{0.4}=490.68 \mathrm{mph}[/latex]

(b) Finally, we check the second derivative to confirm a minimum cost solution:

[latex]\frac{d^{2} \mathrm{C}_{T}}{d v^{2}}=\frac{0.028125}{v^{1 / 2}}+\frac{600,000}{v^{3}} \quad for v>0, and therefore, \frac{d^{2} \mathrm{C}_{T}}{d v^{2}}>0.[/latex]

The company concludes that v = 490.68 mph minimizes the total cost of this particular airplane’s flight.

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