|
Section
3: Word Problems
Area
& fenced Problems
|
|
Problem 1
A
farmer wisher to enclose a field next to a river on 3 sides.
He has 2000 ft. of fencing & does not want a fence on
the adjacent to the river. What
dimensions maximize the area?
- The
quantity to be maximized is area
- A
formula for this quantity is A=x*y
- Since
the total fencing is given by 2000=x +2y, we can solve for y
and substitute into the formula A=x*y
Y
= (2000 – x)/2
(Apply
the area formula then sub new y )®
A= x * [(2000 – x)/2]
A(x)=
1000x – (1/2)x2
Now
we can apply the 2nd derivative test
A’
(x) = 1000 – x
A”(x)
= -1
Set
A’(x) = 0: A’(x)=
1000 – x = 0
Critical
value:
x=1000
Maximum
area:
A(1000) = (1000)(1000) –(1/2)(1000)^2=500,000
2nd
Derivative Test: A”(1000)
= -1, negative implies maximum
·
Thus, a maximum area of 500,000 square feet is enclosed by the
2000 feet of fencing when the dimensions are x = 1000 and y = 500.
**************************************************
|
|
Problem 2
A
3-compartment pen is to be constructed out of 1500 feet of fencing
material. What
dimensions would maximize the area?
·
The
quantity to be maximized is area
·
A
formula for this quantity is
A = x*y
·
Since
the total fencing is given by 1500 = 2x + 4y, we can solve
for y and substitute into the formula A = x*y
Y
= [ (1500 – 2x)/4 ]
A
= x * [ (1500 – 2x)/4 ]
A(x)
= 375 + (1/2) x^2
·
We
can now apply the 2nd Derivative test
A’(x)
= 375 – x
A”(x)
= -1
·
Set
A’(x) = 0 « A’(x) = 375 –
x = 0
·
Critical
Value:
« x
= 375
·
Maximum
area:
A(375) = (375)(375) – (1/2)(375)^2 = 70,321.5
·
2nd
Derivative test: A”(375) = -1, negative implies maximum
y
= [ 1500 – 2(375) ] / 4 = 187.5
·
Conclusion: Thus, a
maximum area of 70,312.5 square feet is enclosed by the 1500 feet
of fencing when the dimensions are x = 375 and y = 187.5.
*******************************************
|
|
Problem 3
A
farmer wishes to enclose 3000 Sq.’ with 6 compartments of equal
area. What dimensions
would minimize the amount of the fencing?
·
The
quantity to be minimized is the length of the fence
·
A
formula for this quantity is L = 3x + 4y.
·
Since
the area is given by 3000 = x*y, we can solve for y and substitute
into the formula L = 3x + 4y
Y
= 3000 / x
L
= 3x + 4( 3000 / x )
L
= 3x + ( 12,000 / x )
- We
can now apply the 2nd derivative test
L
‘(x)= 3 - ( 12,000 / x ^2)
L
“(x)= 24,000 / x
^3)
- Let
L’(x) = 0 «
L ‘(x) = 3 - (
12,000 / x ^2) =0
X^2 = 4000
- Critical
value:
x = about 63.246
- Minimum
fence: L (63.246) = 3(63.246)
+ ( 12,000
/ 63.246 )
= about 379.47
·
2nd
Derivative test: L”(
63.246 ) = 2400 / 63.246 ^3
positive
implies minimum
·
Conclusion: A
minimum of 379.47 ft of fence is required to enclose 3000 sq. ft
with the six equal compartments.
X = 63.246, Y = about 47.434
***************************************
|
|
Problem 4
A
formal flower garden is to cover 800 sq ft and have 2 equally
divided sub areas. What
dimension would minimize the amount of fencing needed?
- The
quantity to be minimized is the length of the fence
- A
formula for this quantity is L = 2x + 3 y
- Since
the area is given by 800 = x*y, we can solve for y and
substitute into the formula L = 2x + 3 y
Y
= 800 / x
L
= 2x + 3 (800 / x )
L(x)
= 2x + (2400 / x )
- We
can apply the 2nd derivative test.
L’(x)
= 2 – 24,000 / x^2
L”(x)
= 4800 / x^3
- Set
L’(x) = 0 «
L’(x) = 2 – 24,000 / x^2
X^2 = 1200
- Critical
value:
X = about 34.64
- Minimum
fence: L (34.64)= 2(34.64) + 3 (800 / 34.64 )
X = about 138.56
- 2nd
Derivative test: L”
(34.64 ) = 4800 / (34.64)^ 3
positive
implies minimum.
- Conclusion:
A minimum of 138.56 ft of fencing is required to
enclose 800 sq ft of flower garden with dimension x = 34.64
and y = 23.09
*********************************************
|
|
Problem 5
A
Rectangular fence costs $12 per ft on sides x but only $4 per foot
on sides y. If the
fence is to enclose 4000 sq ft, what dimensions would minimize the
cost?
- The
quality to be minimized is the cost
- A
formula for this quantity is C = 12*2*x + 4*2*y
- Since
the area is given by 4000 = x * y, we can solve for y and
substitute into the formula C = 12*2*x + 4*2*y
Y
= [ 4000 / x ]
C
= 12*2*x + 4*2*[ 4000 / x ]
C
= 24x + 8[ 4000 / x ]
C(x)
= 24x + 32,000 / x
·
We
can now apply the 2nd derivative test.
C’(x)
= 24 - 32,000 / x^2
C”(x)
= 64,000 / x^3
·
Set
C’ = 0 « C’(x)
= 24 - 32,000 / x^2 =0
X^2 = 32,000/24
X = about 36.515
·
Minimum
cost: C(x) = 24(36.515) + 32,000 /(36.515)
= about 1,752.71
·
2nd derivative test: C”(36.515) = 64,000
/(36.515)^3
Positive implies minimum
·
Conclusion: A minimum cost of $1,752.71 is required to
enclose 4000 sq. ft of yard with dimensions X=(36.515)
and y = 109.54
******************************************
|
Problem 6
The
cost of a garden fence on side x is $6 per foot and $4 per foot on
sides y. No fence is
use next to the building. What
dimensions will minimize cost if the area is 400 sq ft.?
·
The
quantity to be minimized is the cost.
·
A
formula for this quantity is C = 6*x + 4*2*y
·
Since
the area is given by 400=x*y, we can solve for y and substitute
into the formula C = 6x + 8y
Y
= (400/x)
C
= 6x + 8y
C(x)=
6x + 8(400/x)
C(x)=
6x + (3200/x)
- We
can now apply the 2nd derivative test
C’(x)=
6 + (3200/x^2)
C”(x)=
6400/x^3)
C’(x)=
6 + (3200/x^2)=0
X^2
= 3200/6
- Critical
value: x = about 23.09
- Minimum
cost: C(23.09)= 6(23.09) + (3200/23.09)
X =
about 277.13
- 2nd
derivative test: C”
(23.09) = 6400/(23.09)^3),
positive
implies minimum
- Conclusion:
A minimum cost of $277.13 is required to enclose 400 sq ft of
yard
with x = 23.09 and y = 17.32
|
Problem 7
A
homeowner has $1000 to buy fencing for a rectangular dog pen. The cost of fencing on the north side is $8 per foot and $3
per foot on the other sides.
What dimensions would maximize the area?
- The
quantity to be maximized is area
- A
formula for this quantity is A=x*y
- Since
the cost is given by 1000 = 8*x + 3*x + 3*2*y, we can solve
for y and substitute into the formula A=X*Y
1000
= 8*x + 3*x + 3*2*y
1000
= 11x + 6y
Y
= (1000 – 11x)/6
A=x*(1000
– 11x)/6
A(x)=
(500/3)x – (11/6)x^2
- We
can now apply the 2nd derivative test
A’(x)
= 500/3 – 11/3x
A”(x)=
-11/3
- Set
A’(x)= 0 A’(x)
= 500/3 – 11/3x = 0
- Critical
value: x
= 500/11
- Maximum
area: A(500/11)=500/3(500/11)–(11/6)(500/11)^2
x= about 3787.89
- 2nd
Derivative test: A”( 500/11)= -11/3
Negative
implies Maximum
- Conclusion:
A maximum area of 3787.89 occurs when the dimensions are x =
500/11 and Y = 83.333
|
|
Alternate Quiz Problem
|
A farmer has 200
meters of fence to use in fencing in a rectangular
pasture bordering on a straight river. The farmer is not
going to fence along the river so he will only be putting
fence along the three sides shown in green in the picture on
the right. Find the function that gives the total
enclosed area as a function of x and find the dimensions of
the rectangular pasture so fenced that will yield the
maximum enclosed area. Click
here or on the picture to see an animation. In the
animation the graph in red represents the total area divided
by 50 (in order to fit on the screen) as a function of x.
Solution:
- The
quantity to be maximized is area
- A
formula for this quantity is A=x*y
- Since
the total fencing is given by
L =x + 2y, we can solve for y and substitute into
the formula A=x*y
Y = (200 – x)/2
(Apply
the area formula then sub new y )® A= x *
[(200 – x)/2]
A(x)= 100x – (1/2)x2
- We
can now apply the 2nd derivative test
A’(x)=100 –(2)(1/2)x
«
A’(x)=100 – x
A”(x) = -1
- Set
A’(x) = 0:
A’(x) = 100 – x = 0
- Maximum
area: A(100) = (100)(100) –(1/2)(100)^2 = 5,000
2nd
Derivative Test: A”(5000)
= -1, negative implies maximum
- Now
find Y value: Y
= (200 – 100)/2 = 50
- Conclusion:
A
maximum area of 5,000 square feet is enclosed by
the 200 feet of fencing when the dimensions are x = 100
and y = 50.
|
|
|
|
