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NNADVOKAT [17]
2 years ago
6

A field bordering a straight stream is to be enclosed. The side bordering the stream is not to be fenced. If 1000 yards of fenci

ng material is to be used, what are the dimensions of the largest rectangular field that can be fenced? What is the maximum area?
Mathematics
1 answer:
sveta [45]2 years ago
3 0

Answer:

x  =  500 yd

y  =  250 yd

A(max) = 125000 yd²

Step-by-step explanation:

Let´s call x the side parallel to the stream ( only one side to be fenced )

y the other side of the rectangular area

Then the perimeter of the  rectangle  is  p  =  2*x  + 2* y  ( but only 1 x will be fenced)

p  =  x  +  2*y

1000  =  x   +  2 * y         ⇒    y  =  (1000 - x )/ 2

And   A(r)  =  x * y

Are as fuction of x

A(x)  =  x  *  (  1000  -  x ) / 2

A(x)  =   1000*x / 2 -  x² / 2

A´(x)  =  500  -  2*x/2

A´(x)  =  0              500   -  x   =  0

x  =  500 yd

To find out if this value will bring function A to a maximum value we get the second derivative

C´´(x)  =  -1         C´´(x) < 0  then efectevly we got a maximum at  x  = 500

The side  y  = ( 1000  -  x ) / 2

y  =  500/ 2

y  =  250 yd

A(max)  =  250 * 500

A(max) = 125000 yd²

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Let <em>x</em> be the width and <em>y </em>the length of the rectangular field.

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20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30

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