4.63 clark's country pet resort is fencing a new play area for dogs. the manager has purchased 130 yd of fence to enclose a rectangular pen. the area of the pen must be 1050 ydsquared. what are the dimensions of the pen?

Question

Alessandro Taylor

Mathematics

16 November, 04:51

4.63 clark's country pet resort is fencing a new play area for dogs. the manager has purchased 130 yd of fence to enclose a rectangular pen. the area of the pen must be 1050 ydsquared. what are the dimensions of the pen?

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Willis · Answer 1

Let W and L be width and length of the rectangular pen respectively.

Therefore,

Circumference, C = 2W+2L = 130 yd

Area, A = LW = 1050 yd^2=> L = 1050/W

Using the circumference expression and substituting for L;

130 = 2W + 2 (1050/W) = 2W+2100/W

130*W = 2W*W + 2100

130W = 2W^2 + 2100

2W^2-130W+2100 = 0

Solving for W;

W = [ - (-130) + / - Sqrt ((-130) ^2-4 (2) (2100) ]/2*2 = 32.5 + / - 2.5

W = 30 or 35 yd

When W = 30, L = 1050/30 = 35

When W = 35, L = 1050/35 = 30

Therefore, W = 30 yd and L = 35 yd.