Question
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
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
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