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NeX [460]
4 years ago
13

Find the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible enclosed area g

iven 300 feet of fencing. (Assume that the length is greater than or equal to the width.) ft length width ft Tutorial Additional Materials eBook Example Video
Mathematics
1 answer:
Serjik [45]4 years ago
3 0

Answer:Dimensions are of one corral is:

Length=50ft

Width= 37.5ft

Combined area=1875ft

Step-by-step explanation:

Let L= length

Let w = width

There are 3 pieces of fencing: 3L + 4w =300

W = (300 -3L)/ 4 ....equation 1

Combined area of the corrals is given by:

A= L × 2w...eq2

Put eq1 into eq2

A= L ×2(300-3L) /4

A= 1/2 (300 -3 L)^2

A= -(3/2)L^2 + 150

Using -b/2a to solve for L in the quadratic equation

-150/2(3/2) = (150×2)/(2×3)

L = 50ft

W= 300-3(50)/4

W=150/4=37.5ft

Combined are=50×37.5=1875ft

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A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

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A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

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A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
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