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gulaghasi [49]
3 years ago
8

Aidan has 20 ft of fence with which to build a rectangular dog run. What is the maximum area he can enclose? Enter your answer i

n the box.
Mathematics
1 answer:
irina1246 [14]3 years ago
7 0
<h2>Maximum area is 25 m²</h2>

Explanation:

Let L be the length and W be the width.

Aidan has 20 ft of fence with which to build a rectangular dog run.

         Fencing = 2L + 2W  = 20 ft

                     L + W = 10

                      W = 10 - L

We need to find what is the largest area that can be enclosed.

     Area = Length x Width

      A = LW

       A = L x (10-L) = 10 L - L²

For maximum area differential is zero

So we have

      dA = 0

      10 - 2 L = 0

        L = 5 m

      W = 10 - 5 = 5 m

Area = 5 x 5 = 25 m²

Maximum area is 25 m²

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Step-by-step explanation:

1) find the distance of the points

\sqrt{(x2-x1)^2 + (y2 -y1)^2}

AB = \sqrt{(4-2)^2 + (7-5)^2} = \sqrt{4 + 4} = \sqrt{8} = 2 \sqrt{2}

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