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Hoochie [10]
3 years ago
9

For tax reasons, I need to create a rectangular vegetable patch with an area of exactly 50 sq. ft. The fencing for the east and

west sides costs $4 per foot, and the fencing for the north and south sides costs only $2 per foot. What are the dimensions of the vegetable patch with the least expensive fence? ...?
Mathematics
2 answers:
balu736 [363]3 years ago
4 0
Let the dimension of the patch be x for the east and west sides and y for the north and south sides, then
xy = 50 . . . (1)
xdy/dx + y = 0 . . . (2)

Also, Total cost for fencing is 4x + 4x + 2y + 2y = 8x + 4y
For minimum cost, d(8x + 4y)/dx = 0
8 + 4dy/dx = 0
dy/dx = -8/4 = -2 . . . (3)

Putting (3) in (2) gives
-2x + y = 0
y = 2x

Substituting for y in (1) gives
x(2x) = 50
2x^2 = 50
x^2 = 25
x = 5
y = 2(5) = 10

Therefore, for minimum cost, the east and west sides should be 5 feet while the north and south sides should be 10 feet.
Elden [556K]3 years ago
4 0
The solution to the problem is as follows:


East- west length is "w". North-south length is " n". n = 50 / w 
<span>
2*4$ w + 2* 2$ *50/w = cost </span>
<span>
8w +400/w = cost </span>
<span>
To find the minimum take first derivitive and then set it = 0 </span>
<span>
8 - 200 /w ^2 ) = 0 </span>
<span>
8w^2 = 200 </span>
<span>
w = (200/8) ^-2 </span>
<span>
w = 5 feet </span>
<span>
n =50/11 = 4.55 feet </span>


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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