Optimization Problems.

  • Thread starter decamij
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  • #1
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I have couple of questions on optimization, i don't want the answer, i just want to know what i have to do to approach this question. Keep in mind that I am in grade 12 calculus, i.e. PLEASE don't give me some crazy university answer with equations i've never seen before. Anyways, here are the problems:

1)At 11AM, a 747 jet is travelling east at 800km/h. At the same instant, a DC-8 is 45km east and 90km north of the 747. It is at the same altitude travelling south at 600km/h. What is the closest distance of approach of the planes, and at what time does it occur?

(answer is 45km; at 11:05AM)

2)A sailor in a boat 8km off a straight coastline wants to reach a point on shore 10km from the point directly opposite her present position in the shortest possible time. Where should she land the boat and how long does it take her to reach her destination if she can row ar 4km/h and run at 6km/h?

(answer is 7.2 km down the shore;3.15h)

Thanx a bunch!!
 

Answers and Replies

  • #2
ShawnD
Science Advisor
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decamij said:
1)At 11AM, a 747 jet is travelling east at 800km/h. At the same instant, a DC-8 is 45km east and 90km north of the 747. It is at the same altitude travelling south at 600km/h. What is the closest distance of approach of the planes, and at what time does it occur?

(answer is 45km; at 11:05AM)
Damn it, I was doing this problem for like 2 hours before I realized I forgot some fundamental rules of doing derivatives! :yuck:

Anyway, set east/west as X, then north/south as Y. Here is the distance formula:

[tex]D = (X^2 + Y^2)^{\frac{1}{2}}[/tex]

Your X value is (45 - 800t). The 800 is negative because the X gap between the planes is decreasing.
Your Y value is (90 - 600t). The 600 is negative because the Y gap is decreasing.
Now sub that into the original equation

[tex]D = [(45 - 800t)^2 + (90 - 600t)^2]^{\frac{1}{2}}[/tex]

If you want more help, scroll way down. My answer continues on.

























































Now take the derivative.

[tex]\frac{dD}{dt} = \frac{1}{2}[(45 - 800t)^2 + (90-600t)^2]^{\frac{-1}{2}} [2(45 - 800t)(-800) + 2(90-600t)(-600)][/tex]

Remember that dD/dt is 0. If you have a TI calculator, you can type that massive formula into the equation solver and come out with the answer t = 0.09, which is hours. Multiply that by 60 to get 5.4 minutes (11:05AM). Fill that 0.09 back into the original distance formula (first formula in my post) and you'll get 45km. :biggrin:

Sorry for posting the whole answer, but I put way too much work into that problem to just give a hint :wink:
 
Last edited:
  • #3
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Well, about the second question, if the sailor lands the boat at X km down the point directly opposite her then she has rowed [tex]\sqrt{8^{2} + x^{2}}[/tex] km and she needs to run another (10-x) km.
Using this calculate how much time she spends passing each part and then do the derivation.
 
  • #4
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Thanx a bunch
 

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