**Minimizing Costs** Suppose the cost per hour incurred in operating a cruise ship is \(6a + b \frac{2}{v^3}\) dollars per hour, where \(a\) and \(b\) are positive constants and \(v\) is the ship's speed in miles per hour. At what speed (in miles per hour) should the ship be operated between two ports, at a distance \(D\) miles apart, to minimize the cost? (Hint: Minimize the cost, not the cost per hour.)\(v =\) [ 0 ] mphThere are no graphs or diagrams. The problem requires finding the speed that minimizes the total cost for traveling between ports.

Name: **Minimizing Costs** Suppose the cost per hour incurred in operating
Uploaded: 2024-03-20T07:07:26.000Z
Channel: Bartleby
Description: **Minimizing Costs** Suppose the cost per hour incurred in operating a cruise ship is \(6a + b \frac{2}{v^3}\) dollars per hour, where \(a\) and \(b\) are positive constants and \(v\) is the ship's speed in miles per hour. At what speed (in miles p

Given that, the cost per hour incurred in operating a cruise ship is 6a+b2v3 dollars per hour. It is known that, the formula for speed is Speed=Distancetime. Let t be the time taken by the ship to travel from port 1 to port 2 , d represents the travel distance and s is speed of the ship then the time can be calculated as t=ds. From the given information, the total cost incurred is Ct=t6a+b2v2. Ct=t6a+b2dt3 since v=dt=6at+tb2d3t3=6at+b2d3t2Differentiate the function Ct=6at+b2d3t2 with respect to t, C't=ddt6at+b2d3t2=6a-2b2d3t3 Equate the derivative to 0 to find the critical number. 6a-2b2d3t3=06a=2b2d3t3t3=2b2d36at=b2d33a3 , t=-323a23b23d336a±i36a23b23d332at=b2d33a13 , t=-323a23b23d336a±i36a23b23d332a Note that, the travelled time is always positive so the root t=-323a23b23d336a±i36a23b23d332a is not valid. Therefore, the only critical number is t=b2d33a13.Find the second derivative of the function Ct=6at+b2d3t2 as follows, C''t=ddtC't=ddt6a-2b2d3t3=6b2d3x4 Substitute t=b2d33a13in C''t=6b2d3x4, C''t=6b2d3b2d33a134=6b2d3b2d33a43>0 since a>0, b>0 and d>0, t>0 Thus, the second derivative for the critical number is positive, so the function Ct=6at+b2d3t2 attains the minimum at t=b2d33a13.Obtain the value of v. v=dt=db2d33a13=db23a13d313=db23a13d=1b23a13=3ab213=3ab23 Therefore, the speed of ship to minimize the cost incurred is 3ab23.