Solving Optimization Problem Homework

  • Thread starter Thread starter chompysj
  • Start date Start date
  • Tags Tags
    Optimization
Click For Summary
The discussion revolves around solving an optimization problem involving distances traveled on land and water, using the Pythagorean theorem and area calculations. The user has defined variables for distances and attempted to apply the area formula, but is confused about its relevance to the problem. A key point raised is that the focus should be on calculating the total time taken for both distances rather than the area. The solution requires expressing everything in terms of the variable w to find the minimum time. Clarification is provided that the area is not necessary for solving the optimization problem.
chompysj
Messages
19
Reaction score
0

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!
 
Physics news on Phys.org
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?


 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 24 ·
Replies
24
Views
2K
Replies
7
Views
2K
Replies
3
Views
4K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K