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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.

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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.

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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

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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

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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

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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)

  • Set C(x) = 0

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

 

  • Critical value:  x = 100

 

  • 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.