L2 Extreme Values of Distance and Time and Economics - Practice 1
Completion requirements
Unit 5
Applications of Derivatives
A. Maximum and Minimum Problems
Lesson 2: Extreme Values of Distance and Time and Economics
Practice
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Instructions: Click the Download File button to download a printable PDF of the questions. Answer each of the following practice questions on a separate piece of paper. Step by step solutions are provided under the Solutions tab. You will learn the material more thoroughly if you complete the questions before checking the answers.
1.
At «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»10«/mn»«mo»:«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi»am«/mi»«/math», a cargo ship is sailing west at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»20«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math». At the same time, a customs boat «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»60«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» due north of the ship is travelling south at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»30«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math». When will the boats be closest to each other?
2.
A cabin is situated on an island, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» from the nearest point on the mainland. A store on the mainland, where supplies can be purchased, is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»3«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» down the shoreline from that point. If a motor boat travels at a speed of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»4«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math» and the average person can walk «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»6«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math», toward what point on the mainland should the boat be aimed in order to reach the store in the least time?
1.
At «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»10«/mn»«mo»:«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi»am«/mi»«/math», a cargo ship is sailing west at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»20«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math». At the same time, a customs boat «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»60«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» due north of the ship is travelling south at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»30«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math». When will the boats be closest to each other?
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» be the time travelled.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»z«/mi»«/math» be the distance to be minimized.
After a period of time «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math», in hours, the cargo ship travels a distance of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»20«/mn»«mi»t«/mi»«mo»§#160;«/mo»«mi»km«/mi»«/math» and the customs boat travels a distance of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»60«/mn»«mo»-«/mo»«mn»30«/mn»«mi»t«/mi»«mo»§#160;«/mo»«mi»km«/mi»«/math». The new positions of the boats are shown in the diagram.
Use the Pythagorean Theorem to find the distance to be minimized.
Determine the derivative of the distance function and equate it to zero.
To verify if the critical point «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«/math» is a minimum, consider the following:
If the boats sail for an hour, they would be approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»36«/mn»«mo».«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» apart.
If the boats sail for «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«/math» hour, they would be approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»33«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» apart.
Since «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msqrt»«mn»1300«/mn»«/msqrt»«mo»§#62;«/mo»«msqrt»«mfrac»«mrow»«mn»187«/mn»«mi mathvariant=¨normal¨» «/mi»«mn»200«/mn»«/mrow»«mn»169«/mn»«/mfrac»«/msqrt»«/math» the critical point of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«/math» must be a minimum.
Therefore, the boats are closest together at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»h«/mi»«/math» (approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»1«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»38«/mn»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»h«/mi»«/math»).
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»z«/mi»«/math» be the distance to be minimized.
After a period of time «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math», in hours, the cargo ship travels a distance of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»20«/mn»«mi»t«/mi»«mo»§#160;«/mo»«mi»km«/mi»«/math» and the customs boat travels a distance of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»60«/mn»«mo»-«/mo»«mn»30«/mn»«mi»t«/mi»«mo»§#160;«/mo»«mi»km«/mi»«/math». The new positions of the boats are shown in the diagram.
Use the Pythagorean Theorem to find the distance to be minimized.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«msup»«mi»z«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mfenced»«mrow»«mn»60«/mn»«mo»§#8722;«/mo»«mn»30«/mn»«mi»t«/mi»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«mo»+«/mo»«msup»«mfenced»«mrow»«mn»20«/mn»«mi»t«/mi»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo»§#160;«/mo»«mn»600«/mn»«mo»§#8722;«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mn»600«/mn»«mi»t«/mi»«mo»+«/mo»«mn»900«/mn»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»400«/mn»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»300«/mn»«msup»«mi»t«/mi»«mn»2«/mn»«/msup»«mo»§#8722;«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mn»600«/mn»«mi»t«/mi»«mo»+«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mn»600«/mn»«/mtd»«/mtr»«/mtable»«/math»
Determine the derivative of the distance function and equate it to zero.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mn»2«/mn»«mi»z«/mi»«mfrac»«mrow»«mi»d«/mi»«mi»z«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#160;«/mo»«mn»600«/mn»«mi»t«/mi»«mo»§#8722;«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mn»600«/mn»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»z«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»2«/mn»«mo»§#160;«/mo»«mn»600«/mn»«mi»t«/mi»«mo»§#8722;«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mn»600«/mn»«/mrow»«mrow»«mn»2«/mn»«mi»z«/mi»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»300«/mn»«mi»t«/mi»«mo»§#8722;«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mn»800«/mn»«/mrow»«mi»z«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd/»«mtd/»«/mtr»«mtr»«mtd»«mn»0«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»300«/mn»«mi»t«/mi»«mo»§#8722;«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mn»800«/mn»«/mrow»«mi»z«/mi»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mn»0«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»300«/mn»«mi»t«/mi»«mo»§#8722;«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mn»800«/mn»«/mtd»«/mtr»«mtr»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»800«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»300«/mn»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»800«/mn»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»300«/mn»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»t«/mi»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»t«/mi»«/mtd»«/mtr»«/mtable»«/math»
To verify if the critical point «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«/math» is a minimum, consider the following:
- Intuitively try to determine whether the critical point is a minimum value. The boats are sailing away from each other, so there should be no maximum.
- Since there is only one critical point, compare it to another value.
If the boats sail for an hour, they would be approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»36«/mn»«mo».«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» apart.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«msup»«mi»z«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi mathvariant=¨normal¨»(«/mi»«mn»60«/mn»«mo»§#8722;«/mo»«mn»30«/mn»«mi mathvariant=¨normal¨»(«/mi»«mn»1«/mn»«msup»«mi»))«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«msup»«mfenced»«mrow»«mn»20«/mn»«mi mathvariant=¨normal¨»(«/mi»«mn»1«/mn»«mi mathvariant=¨normal¨»)«/mi»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mn»30«/mn»«mn»2«/mn»«/msup»«mo»+«/mo»«msup»«mn»20«/mn»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»900«/mn»«mo»+«/mo»«mn»400«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«mo»§#160;«/mo»«mn»300«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»z«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msqrt»«mn»1«/mn»«mo»§#160;«/mo»«mn»300«/mn»«/msqrt»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»§#8784;«/mo»«/mtd»«mtd»«mn»36«/mn»«mo».«/mo»«mn»1«/mn»«/mtd»«/mtr»«/mtable»«/math»
If the boats sail for «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«/math» hour, they would be approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»33«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» apart.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«msup»«mi»z«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mfenced»«mrow»«mn»60«/mn»«mo»§#8722;«/mo»«mn»30«/mn»«mfenced»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«/mfenced»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«mo»+«/mo»«msup»«mfenced»«mrow»«mn»20«/mn»«mfenced»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«/mfenced»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mfenced»«mfrac»«mn»240«/mn»«mn»13«/mn»«/mfrac»«/mfenced»«mn»2«/mn»«/msup»«mo»+«/mo»«msup»«mfenced»«mfrac»«mn»360«/mn»«mn»13«/mn»«/mfrac»«/mfenced»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»187«/mn»«mi mathvariant=¨normal¨» «/mi»«mn»200«/mn»«/mrow»«mn»169«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»z«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msqrt»«mfrac»«mrow»«mn»187«/mn»«mi mathvariant=¨normal¨» «/mi»«mn»200«/mn»«/mrow»«mn»169«/mn»«/mfrac»«/msqrt»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»§#8784;«/mo»«/mtd»«mtd»«mn»33«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»3«/mn»«/mtd»«/mtr»«/mtable»«/math»
Since «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msqrt»«mn»1300«/mn»«/msqrt»«mo»§#62;«/mo»«msqrt»«mfrac»«mrow»«mn»187«/mn»«mi mathvariant=¨normal¨» «/mi»«mn»200«/mn»«/mrow»«mn»169«/mn»«/mfrac»«/msqrt»«/math» the critical point of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«/math» must be a minimum.
Therefore, the boats are closest together at «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mfrac»«mn»18«/mn»«mn»13«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»h«/mi»«/math» (approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»1«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»38«/mn»«mi mathvariant=¨normal¨» «/mi»«mi mathvariant=¨normal¨»h«/mi»«/math»).
2.
A cabin is situated on an island, «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»2«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» from the nearest point on the mainland. A store on the mainland, where supplies can be purchased, is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»3«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math» down the shoreline from that point. If a motor boat travels at a speed of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»4«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math» and the average person can walk «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»6«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»h«/mi»«/math», toward what point on the mainland should the boat be aimed in order to reach the store in the least time?
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«/math» be the distance along the shoreline to the point where the boat is headed.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» be the total time to be minimized.
The total time it takes to reach the store is as follows.
To find time, use the formula «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mfrac»«mi»d«/mi»«mi»v«/mi»«/mfrac»«/math».
Use Pythagorean Theorem to find the distance the boat travels.
The walking distance can be read from the diagram, that is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«mo»=«/mo»«mn»3«/mn»«mo»§#8722;«/mo»«mi»x«/mi»«/math».
Now, find the time it takes to boat and walk.
Determine the derivative of the time function and equate it to zero.
Solve for «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«/math» where «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«/math».
To find the absolute minimum of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» on the interval «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfenced open=¨[¨ close=¨]¨»«mrow»«mn»0«/mn»«mi mathvariant=¨normal¨»,«/mi»«mo»§#160;«/mo»«mn»3«/mn»«/mrow»«/mfenced»«/math», we evaluate «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» at the critical number «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«mo»§#8801;«/mo»«mfrac»«mrow»«mn»15«/mn»«mo»§#8722;«/mo»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»§#8784;«/mo»«mn»1«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»21«/mn»«/math» and the endpoints.
The absolute minimum time occurs when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«mo»=«/mo»«mn»1«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»21«/mn»«/math».
To minimize the time taken to get to the store, the boat should be aimed «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»km«/mi»«/math» (approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»1«/mn»«mo».«/mo»«mn»8«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math») down the shore from a point directly across from the island and the remaining «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»15«/mn»«mo»§#8722;«/mo»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi»km«/mi»«/math» (approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»1«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math») should be walked.
Let «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» be the total time to be minimized.
The total time it takes to reach the store is as follows.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»t«/mi»«mi»total«/mi»«/msub»«mo»=«/mo»«msub»«mi»t«/mi»«mi»boating«/mi»«/msub»«mo»+«/mo»«msub»«mi»t«/mi»«mi»walking«/mi»«/msub»«/math»
To find time, use the formula «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«mo»=«/mo»«mfrac»«mi»d«/mi»«mi»v«/mi»«/mfrac»«/math».
Use Pythagorean Theorem to find the distance the boat travels.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»d«/mi»«mi»b«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«msup»«mn»2«/mn»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»d«/mi»«mi»b«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msqrt»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/msqrt»«/mtd»«/mtr»«/mtable»«/math»
The walking distance can be read from the diagram, that is «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«mo»=«/mo»«mn»3«/mn»«mo»§#8722;«/mo»«mi»x«/mi»«/math».
Now, find the time it takes to boat and walk.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left center center right center left¨»«mtr»«mtd»«msub»«mi»t«/mi»«mi»b«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«msub»«mi»d«/mi»«mi»b«/mi»«/msub»«msub»«mi»v«/mi»«mi»b«/mi»«/msub»«/mfrac»«/mtd»«mtd/»«mtd/»«mtd»«msub»«mi»t«/mi»«mi»w«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«msub»«mi»v«/mi»«mi»w«/mi»«/msub»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«msqrt»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/msqrt»«mn»4«/mn»«/mfrac»«/mtd»«mtd/»«mtd/»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»3«/mn»«mo»§#8722;«/mo»«mi»x«/mi»«/mrow»«mn»6«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»t«/mi»«mi»total«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msub»«mi»t«/mi»«mi»boating«/mi»«/msub»«mo»+«/mo»«msub»«mi»t«/mi»«mi»walking«/mi»«/msub»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«msqrt»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/msqrt»«mn»4«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mrow»«mn»3«/mn»«mo»§#8722;«/mo»«mi»x«/mi»«/mrow»«mn»6«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»4«/mn»«/mfrac»«msup»«mfenced»«mrow»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/mrow»«/mfenced»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/msup»«mo»+«/mo»«mfenced»«mrow»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»6«/mn»«/mfrac»«mi»x«/mi»«/mrow»«/mfenced»«/mtd»«/mtr»«/mtable»«/math»
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»t«/mi»«mi»total«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msub»«mi»t«/mi»«mi»boating«/mi»«/msub»«mo»+«/mo»«msub»«mi»t«/mi»«mi»walking«/mi»«/msub»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«msqrt»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/msqrt»«mn»4«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mrow»«mn»3«/mn»«mo»§#8722;«/mo»«mi»x«/mi»«/mrow»«mn»6«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»4«/mn»«/mfrac»«msup»«mfenced»«mrow»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/mrow»«/mfenced»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/msup»«mo»+«/mo»«mfenced»«mrow»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»6«/mn»«/mfrac»«mi»x«/mi»«/mrow»«/mfenced»«/mtd»«/mtr»«/mtable»«/math»
Determine the derivative of the time function and equate it to zero.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mfrac»«mrow»«mi»d«/mi»«mi»t«/mi»«/mrow»«mrow»«mi»d«/mi»«mi»x«/mi»«/mrow»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»4«/mn»«/mfrac»«mo»§#8729;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«msup»«mfenced»«mrow»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/mrow»«/mfenced»«mrow»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mrow»«/msup»«mfenced»«mrow»«mn»2«/mn»«mi»x«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»6«/mn»«/mfrac»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»x«/mi»«mrow»«mn»4«/mn»«msqrt»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/msqrt»«/mrow»«/mfrac»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»6«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd/»«mtd/»«/mtr»«mtr»«mtd»«mn»0«/mn»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»x«/mi»«mrow»«mn»4«/mn»«msqrt»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/msqrt»«/mrow»«/mfrac»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»6«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mn»1«/mn»«mn»6«/mn»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mi»x«/mi»«mrow»«mn»4«/mn»«msqrt»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/msqrt»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«msup»«mfenced»«mfrac»«mn»1«/mn»«mn»6«/mn»«/mfrac»«/mfenced»«mn mathcolor=¨#FF0000¨»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msup»«mfenced»«mfrac»«mi»x«/mi»«mrow»«mn»4«/mn»«msqrt»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/msqrt»«/mrow»«/mfrac»«/mfenced»«mn mathcolor=¨#FF0000¨»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mn»1«/mn»«mn»36«/mn»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mrow»«mn»16«/mn»«mfenced»«mrow»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/mrow»«/mfenced»«/mrow»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mn»36«/mn»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»16«/mn»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»64«/mn»«/mtd»«/mtr»«mtr»«mtd»«mn»20«/mn»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»64«/mn»«/mtd»«/mtr»«mtr»«mtd»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»64«/mn»«mn»20«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«msqrt»«mfrac»«mn»64«/mn»«mn»20«/mn»«/mfrac»«/msqrt»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«msqrt»«mfrac»«mn»16«/mn»«mn»5«/mn»«/mfrac»«/msqrt»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«msqrt»«mn»16«/mn»«/msqrt»«msqrt»«mn»5«/mn»«/msqrt»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
Solve for «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«/math» where «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»x«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«/math».
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo»§#8722;«/mo»«mi»x«/mi»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo»§#8722;«/mo»«mfrac»«mrow»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»15«/mn»«mo»§#8722;«/mo»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/math»
To find the absolute minimum of «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» on the interval «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfenced open=¨[¨ close=¨]¨»«mrow»«mn»0«/mn»«mi mathvariant=¨normal¨»,«/mi»«mo»§#160;«/mo»«mn»3«/mn»«/mrow»«/mfenced»«/math», we evaluate «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»t«/mi»«/math» at the critical number «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«mo»§#8801;«/mo»«mfrac»«mrow»«mn»15«/mn»«mo»§#8722;«/mo»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»§#8784;«/mo»«mn»1«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»21«/mn»«/math» and the endpoints.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»t«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«msqrt»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«mo»+«/mo»«mn»4«/mn»«/msqrt»«mn»4«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mrow»«mn»3«/mn»«mo»§#8722;«/mo»«mi»x«/mi»«/mrow»«mn»6«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»t«/mi»«mi mathvariant=¨normal¨»(«/mi»«mn»0«/mn»«mi mathvariant=¨normal¨»)«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»2«/mn»«mn»4«/mn»«/mfrac»«mo»+«/mo»«mfrac»«mn»3«/mn»«mn»6«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»1«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»t«/mi»«mi mathvariant=¨normal¨»(«/mi»«mn»1«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»21«/mn»«mi mathvariant=¨normal¨»)«/mi»«/mtd»«mtd»«mo»§#8784;«/mo»«/mtd»«mtd»«mn»0«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»58«/mn»«mo»+«/mo»«mn»0«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»30«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»§#8784;«/mo»«/mtd»«mtd»«mn»0«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»88«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»t«/mi»«mi mathvariant=¨normal¨»(«/mi»«mn»3«/mn»«mi mathvariant=¨normal¨»)«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«msqrt»«mn»13«/mn»«/msqrt»«mn»4«/mn»«/mfrac»«mo»+«/mo»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd/»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»90«/mn»«/mtd»«/mtr»«/mtable»«/math»
The absolute minimum time occurs when «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«msub»«mi»d«/mi»«mi»w«/mi»«/msub»«mo»=«/mo»«mn»1«/mn»«mi mathvariant=¨normal¨».«/mi»«mn»21«/mn»«/math».
To minimize the time taken to get to the store, the boat should be aimed «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«mo»§#160;«/mo»«mi mathvariant=¨normal¨» «/mi»«mi»km«/mi»«/math» (approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»1«/mn»«mo».«/mo»«mn»8«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math») down the shore from a point directly across from the island and the remaining «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»15«/mn»«mo»§#8722;«/mo»«mn»4«/mn»«msqrt»«mn»5«/mn»«/msqrt»«/mrow»«mn»5«/mn»«/mfrac»«mi mathvariant=¨normal¨» «/mi»«mo»§#160;«/mo»«mi»km«/mi»«/math» (approximately «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn»1«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi»km«/mi»«/math») should be walked.