Solving Optimization Problem Homework

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SUMMARY

The discussion centers on solving an optimization problem involving a man traveling to a shoreline. The key equations utilized are the Pythagorean theorem, represented as 100^2 + w^2 = y^2 and (300-w)^2 + 65^2 = x^2, where 'w' denotes the distance from the left at which the man lands on the shore. The derivatives of the distances traveled are given as dy/dt = 70 yd/min and dx/dt = 120 yd/min. The main conclusion emphasizes that the area of the triangle is irrelevant; instead, the focus should be on minimizing the total time taken to travel both distances.

PREREQUISITES
  • Understanding of the Pythagorean theorem
  • Basic knowledge of derivatives and rates of change
  • Familiarity with optimization techniques in calculus
  • Ability to set up and solve equations involving multiple variables
NEXT STEPS
  • Study optimization problems in calculus, focusing on minimizing functions
  • Learn how to apply the Pythagorean theorem in real-world scenarios
  • Explore the concept of related rates in calculus
  • Practice setting up and solving multi-variable equations
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Students studying calculus, particularly those focusing on optimization problems, as well as educators looking for examples of real-world applications of mathematical concepts.

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Homework Statement



879lnva.jpg


Homework Equations



a^2 + b^2 = c^2
A= (1/2)bh

The Attempt at a Solution



1)I labeled the distance traveled on the ocean as y and the distance traveled on land as x.

2)This one's kinda hard to describe: you know how the 100 yd distance and the shoreline form a triangle? I labeled the bottom of that triangle as w.

3) Pythagorean theorem:

100^2 + w^2 = y^2
(300-w)^2 + 65^2 = x^2

Area of triangle:

(300)(65) - (1/2)(300-w)(65) = 65w + (1/2)(65)(300-w)

Also, the derivatives:

dy/dt = 70 yd/min
dx/dt = 120 yd/min

This is kind of where I'm stuck. Am I on the right track with the whole w thing? Or am I headed in a completely wrong direction? Please help!
 
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chompysj said:

Homework Statement



879lnva.jpg


Homework Equations



a^2 + b^2 = c^2
A= (1/2)bh

The Attempt at a Solution



1)I labeled the distance traveled on the ocean as y and the distance traveled on land as x.

2)This one's kinda hard to describe: you know how the 100 yd distance and the shoreline form a triangle? I labeled the bottom of that triangle as w.
Okay, so "w" is the distance from the left at which the man lands on the shore.

3) Pythagorean theorem:

100^2 + w^2 = y^2
(300-w)^2 + 65^2 = x^2

Area of triangle:

(300)(65) - (1/2)(300-w)(65) = 65w + (1/2)(65)(300-w)
Area of triangle? What in the world does area have to do with anything?

so, the derivatives:

dy/dt = 70 yd/min
dx/dt = 120 yd/min

This is kind of where I'm stuck. Am I on the right track with the whole w thing? Or am I headed in a completely wrong direction? Please help!
"Area" has nothing to do with this problem. If he goes distance y yds at 70 yd/min, how long does it take him? If he goes distance x yds at 120 yd/min, how long does it take him? What is the total time? Now use your formula to put everything in terms of your original variable, w. For what value of w is that time a minimum?


 

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