
Bradford T. answered 03/23/22
Retired Engineer / Upper level math instructor
Let x = the length of the brick wall side and the length of the fence parallel to the brick wall. Let y be the length of each of the other 2 metal fence sides perpendicular to the brick side.
Cost = 30(brick length) + 10(parallel fence) + 10 (the two perpendicular parts) = 30x + 10x + 10(2y)
= 40x + 20y
Area, A = xy = 148 square feet
y = 148/x
C(x) = 40x + 20(148/x)
To find the minimum, find C'(x), set that to zero and solve for x.
C'(x) = 40-20(148/x2)
40 -20(148/x2) = 0
40 = 20(148/x2)
x2 = 148/2 =74
x = ±√74 Use the positive value since this is a length
y = 148/√74 = 2√74
Length of side with bricks (and opposite side) √74 ft.
Length of sides perpendicular to the brick wall 2√74 ft.

Bradford T.
The 3rd fence side is parallel to the brick side.03/23/22
Mariam A.
Why it is 2 metal fence while in the question its 3 sides?03/23/22