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Katena32 [7]
3 years ago
12

A rectangular field is to be fenced off, and then divided into two by a fence running parallel to one of the sides. If 744 meter

s of fencing can be used, find the maximum area that can be enclosed by the fence.\

Physics
2 answers:
Mrac [35]3 years ago
7 0

Explanation:

Below is an attachment containing the solution.

alexandr402 [8]3 years ago
7 0

Answer:

Maximum area enclosed by the fence= 23,064 m²

Explanation:

Size of the fence used = 744m

Let the width of the fence be y

If the fence is divided into two by a fence running in parallel to one side of the fence, then the length, L = (744 - 3y)/2

Area enclosed by the fence, A = Length * Breadth

A = (744 - 3y)/2  *  y

A = 372y - 1.5y²

At maximum area, A'(y) = 0

A'(y) = 372 - 3y

372 - 3y = 0

y = 372/3

y = 124 meters

Maximum area enclosed by the fence = 372(124) - 1.5*124²

A = 46128 - 23064

Maximum area = 23,064 m²

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Answer:

Explanation:

Given

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similarly B travels with 20 mph and in 2 hours

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Thus distance between A and B  is

r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}

|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}

|r_{AB}|=103.45 miles

Velocity of A

v_A=-40sin35\hat{i}+40cos35\hat{j}

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v_B=20sin80\hat{i}+20cos80\hat{j}

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4 0
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3 0
1 year ago
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