Please Answer the assignment question in similar steps as it was shown in the sample question

Assignment:

Complete the following applied optimization word problem. Include a

well-labeled figure, clearly defined objective function of one variable, interval for which that

variable is logical, and supporting mathematical statements to justify that the optimal result

is achieved for the function on the interval within a grammatically correct exposition.

A hiker is returning to a campsite that is 5 miles east and 2 miles north from their current

position. The hiker is currently walking east along a path at a pace of 3 miles per hour.

The hiker plans to leave the path and walk in a straight line diagonally towards the cabin

through the woods. The hiker can only walk at a pace of 1 mile per hour through the woods.

Where should the hiker leave the path and turn into the woods if they want to get to the

campsite as quickly as possible?

Sample Problem and Solution:

Problem:

A farmer wants to build a fenced enclosure on his farmland for his free-

range ostriches. He wants a rectangular ostrich pen built with 1000 feet of fencing

material to be divided into three equal sections by two interior fences that run parallel

to a pair of exterior side fences. Determine how to set up the ostrich pen so that the

maximum possible area is enclosed, and find this maximum area.

Answer:

The rectangular pen should be set up so that the sides parallel to the interior

fences are each 125 feet long and the other pair of sides are each 250 feet long. This

would provide an enclosed area of 31,250 square feet.

1

Solution:

The rectangular fence that we are considering is given in the following

figure, where we let

x

represent the length of each side in the pair that is not parallel

to the interior fences and we let

y

represent the length of each side in the pair that is

parallel to the interior fences as well as the length of the two interior fences.

y

x

Since we want to maximize the area of the pen, the objective function for this opti-

mization will represent the total area of the pen, given by

A

=

xy.

Since this objective function is currently in terms of two variables, we must use addi-

tional information given in the problem to help us eliminate one of the two variables.

We are given that we have 1000 feet of fencing to build the pen, so the constraint

equation will set this total amount of fencing equal to the sum of all of the lengths of

the pieces of fence that are to be constructed. So, we have the constraint equation

1000 = 2

x

+ 4

y.

We first notice that both sides of the constraint equation have the common factor of

2, so we divide each side of the equation by 2 to get

500 =

x

+ 2

y.

Since the coefficient in front of the

x

is 1, we can avoid introducing fractions by solving

the constraint equation for

x

to get

x

= 500

2

y.

Then we substitute the expression 500

2

y

into the variable

x

in the objective function,

giving us the objective function in one variable

A

= (500

2

y

)

y.

In order simplify the derivative a bit, we go ahead and distribute the

y

across the two

terms in parenthesis to get the objective function

A

= 500

y

2

y

2

.

Next, we use the constraint equation and the fact that we must have sides of positive

length to determine the interval over which we must optimize. We start with the

2

fact that

y >

0 and use the fact that

x >

0 to determine the upper bound on the

feasible domain. Since the constraint equation gives us that

x

= 500

2

y

, we solve the

inequality

500

2

y >

0

to get

y <

250, giving us the interval of

(0

,

250)

for our variable

y

. We are now ready to find the global max of of the objective function

on the interval that follows:

A

= 500

y

2

y

2

(0

,

250)

We first notice that

A

does not have any discontinuities on the interval, so we move to

identifying critical points. To solve the problem, we take the derivative of

A

to get

A

= 500

4

y

and notice that it is defined everywhere. So, we set the derivative equal to zero and

solve for

y

:

0 = 500

4

y

4

y

= 500

y

= 125

The critical point of

y

= 125 does fall in the interval (0

,

250), and it is the only critical

point in the interval. So, to justify that this critical point is the location of the desired

global maximum of the objective function

A

, we only need to establish that it is the

location of a local maximum. Since

A

=

4

is negative always, it is certainly negative at the critical point. Thus, we do have a

local, and hence global, maximum of the objective function

A

at the critical point

y

= 125

.

To get the value of

x

, we plug

y

= 125 back into the constraint equation and get that

x

= 500

2(125) = 250

.

Utilizing the values of both

x

= 250 and

y

= 125, we compute that the maximum area

is

A

= (250)(125) = 31

,

250

.

So, our conclusion is that the rectangular pen should be set up so that the sides parallel

to the interior fences are each 125 feet long and the other pair of sides are each 250

feet long. This would provide an enclosed area of 31,250 square feet