Optimization: A Distance Example - Maple Programming Help

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Optimization: A Distance Example

Main Concept

An optimization problem involves finding the best solution from all feasible solutions. One is usually solving for the largest or smallest value of a function, such as the shortest distance or the largest volume.

A minimum or maximum of a continuous function over a range must occur either at one of the endpoints of the range, or at a point where the derivative of the function is 0 (and thus the tangent line is horizontal). These are called critical points.

Steps

1. 

Identify what value is to be maximized or minimized.

2. 

Define the constraints.

3. 

Draw a sketch or a diagram of the problem.

4. 

Identify the quantity that can be adjusted, called the variable, and give it a name, such as x.

5. 

Write down a function expressing the value to be optimized in terms of x.

6. 

Differentiate the equation with respect to x.

7. 

Set the equation to 0 and solve for x.

8. 

Check the value of the function at the end points.

 

Problem: Every morning Tom leaves his house, gets water from the river, and takes it to the farm. What is the shortest possible path that Tom has to walk?

 

Let x be the distance downstream from the house at the point where Tom gets water from the river.

Adjust the value of x using the slider to find value that minimizes the distance traveled.

x =

Numerical solution

d = x2 +102 , D= 30x2 + 202

x =  0x 30, x ℝ

Total distance T:

T

=

 d+ D 

=

x2+102 + 30x2+202

Calculate first derivative:

ⅆTⅆ x

=

 xx2+102 30x30x2+202

Set it to 0:

0

=

  xx2+102 30x30x2+202

Solve for an x value that minimizes T:

0

=

 x 30x2+202 30xx2+102

30xx2+102  

=

x 30x2+202

30x2x2+1022 

=

 x2 30x2+2022

30x2x2+102

=

x2 30x2+202

x2   60 x + 900x2+100

=

 x2 x2   60 x + 900+400

 x460x3+1000x26000 x+90000

=

x4   60 x3 + 1300 x2

300x26000 x +90000

=

0

300x+30x10

=

0 

Choose the positive root:

x

=

10

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