Question 773758
Randy is building a rectangular, fenced dog run beside his barn.
 He has a budget of $1050 for fencing and it costs him $8.75 per meter for the materials.
 He plans to use the side of the barn as one side of the fenced area.
 What are the dimensions of a dog run that maximizes the area Randy can enclose?
:
Find out how much fence he can get with $1050,
 this is the total distance of the three sides
:
1050/8.75 = 120 meters
therefore 
L + 2W = 120
L = (120-2W)
:
Area
A = L*W
Replace L with (120-2W)
A = W(120-2W)
A = -2W^2 + 120W; a quadratic equation
We can find the max area by finding the axis of symmetry x = -b/(2a)
In this equation we have
W = {{{(-120)/(2*-2)}}}
W = 30 meters width for max area
L = 120-2(30)
L = 60 meters length for max area
therefore the dimension: 60 by 30 meters, that would be 1800 sq/m