A farmer wishes to create a rectangular enclosure of ABCD, of area 525m^2, as shown below.
D________________C
 
 
 
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!!
D________________C
 
 
 
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!!

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.
.. 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.

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
(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
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