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Help with an area of a rectangle math problem?!
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# Help with an area of a rectangle math problem?!

[From: ] [author: ] [Date: 11-05-02] [Hit: ]
.The minimum cost is \$420.(1)..since its a rectangle........
A farmer wishes to create a rectangular enclosure of ABCD, of area 525m^2, as shown below.
D________________C
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A|_______________|B
The fencing used for side AB costs \$11 per meter. The fencing for the other three sides costs \$3 per meter. The farmer creates an enclosure so that the cost is a minimum. Find this minimum cost.

Thank you so much for your help!!

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The cost as a function of the length of side AB is given by
.. c = 11*AB + 3(525/AB) + 3*AB + 3(525/AB)
.. c = 14*AB+3150/AB

The derivative of this with respect to AB is
.. c' = 14 - 3150/AB^2

This will be zero when
.. 14 = 3150/AB^2
.. AB = sqrt(3150/14) = 15

The minimum cost will be
.. c = 14*15 + 3150/15 = 420

The minimum cost is \$420.

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area = L x w = AB x AD
(1).. AB x AD = 525 m²

cost = 11AB + 3 AD + 3 DC + 3 CB

since it's a rectangle..length AB = length CB and since length AD = length DC... meaning...
(2).. cost = 14 AB + 6 AD

rearranging and substituting (1) into (2)
cost = 14 x (525 / AD) + 6 AD

*********
and that is like
y = 7350 / x + 6 x

and the derivative is..
y' = -7350/x² + 6

and the mins and max's occur when y' = 0
0 = -7350/x² + 6
6 = -7350/x²
x² = -7350/6
x = +/- 35

it doesn't make sense to have a - length.. so x = 35.. ie.. AD = 35m
and that means.. AB = 525 / 35 = 15

*********
ie..
Length AB = length DC = 15m
Length AD = length BC = 35m

-
Let AB = CD = x

Then AD = BC = 525/x

cost = y = 11x + 3(x + 525/x + 525/x) = 14 x+3150/x

dy/dx = 14 - 3150/x^2

dy/dx = 0
14 - 3150/x^2 = 0
14x^2 = 3150
x = 15 or x = -15
x = 15

cost = (14 x+3150/x where x = 15) = \$270
1
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