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A city park commission received a donation of playground equipment from a parents' organization. The area of the playground needs to be 256 square yards for the children to use it safely. The playground will be rectangular. The city will also put a fence around the playground. The perimeter, P, of the fence includes the gates. To save money, the city wants the least perimeter of fencing for the area of 256 square yards. With one side 8 yards longer than the other side, what are the side lengths for the least perimeter of fencing?

Guest Sep 6, 2017

#1**+2 **

Playground

256 yd^2

P least perimeter

1.

All except b satisfy the model. However from the wording of the playground c is probably the most appropriate.

2.

Step 7. It should have read x(1/2*P - x) = 256 though I don't see what it proves.

3.

x(x + 8) = 256

x^2 + 8x - 256 = 0

By quadratic formula

x = [-b ± √(b^2 - 4ac)]/(2a)

= [-8 ± √(8^2 + 4*256)]/2

= (-8 ± √1088)/2

= [-8 ± √(64*17)]/2

= (-8 ± 8√17)/2

= -4 ± 4√17 ….. minus in ± does not apply in this case

x = -4 + 4√17

≈ 12.49

x + 8 = 20.49

_12.49_ yards and _20.49_yards

4.

x^2 = 256

x = √256

= 16

_16_ yards and _16_yards

5.

A = x(P/2 - x)

= -x^2 + Px/2

Taking the derivative

A' = -2x + P/2

Setting it to zero

-2x + P/2 = 0

x = P/4

So

A = (P/4)(P/2 - P/4)

= (P/4)^2

Therefore it is a square

256 = (P/4)^2

P/4 = √256

= 16

P = 4*16

= 64

The least fencing is 64 yards

im pretty darn sure im correct

cowgirlkatie03 Sep 6, 2017

#1**+2 **

Best Answer

Playground

256 yd^2

P least perimeter

1.

All except b satisfy the model. However from the wording of the playground c is probably the most appropriate.

2.

Step 7. It should have read x(1/2*P - x) = 256 though I don't see what it proves.

3.

x(x + 8) = 256

x^2 + 8x - 256 = 0

By quadratic formula

x = [-b ± √(b^2 - 4ac)]/(2a)

= [-8 ± √(8^2 + 4*256)]/2

= (-8 ± √1088)/2

= [-8 ± √(64*17)]/2

= (-8 ± 8√17)/2

= -4 ± 4√17 ….. minus in ± does not apply in this case

x = -4 + 4√17

≈ 12.49

x + 8 = 20.49

_12.49_ yards and _20.49_yards

4.

x^2 = 256

x = √256

= 16

_16_ yards and _16_yards

5.

A = x(P/2 - x)

= -x^2 + Px/2

Taking the derivative

A' = -2x + P/2

Setting it to zero

-2x + P/2 = 0

x = P/4

So

A = (P/4)(P/2 - P/4)

= (P/4)^2

Therefore it is a square

256 = (P/4)^2

P/4 = √256

= 16

P = 4*16

= 64

The least fencing is 64 yards

im pretty darn sure im correct

cowgirlkatie03 Sep 6, 2017