Aregular garden of area of 75 ft^2 is bounded on three sides by a wall costing $8 per ft and on the fourth by a fence costing $4 per ft. what are the most economical dimensions of the garden?

It is a problem of maximums and minimums.
Let´s:
x=large
y=width

xy=75   ⇒  y=75/x

Lets;
C(x,y)=cost of the fence and wall
C(x,y)=8(x+2y)+4x
C(x,y)=8x+16y+4x
C(x,y)=12x+16y
C(x)=12x+16(75/x)
C(x)=(12x²+1200)/x

1) we have to do the first derivative:
C´(x)=[x(24x)-(12x²+1200)] / x²
C´(x)=(24x²-12x²-1200) / x²
C´(x)=(12x²-1200) / x²

2) we get values when C´(x)=0
C´(x)=0
(12x²-1200)/x² =0
12x²-1200=0
x=⁺₋√(1200/12)
x₁=10
x₂=-10   (this solution is not valid).

3)we have to do the second derivative:
C´´(x)=2400/x³
C´´(10)=2400/(10³)>0 therefore, we have a minimun when x=10

4) we find out the dimensions of the garden:
x=10
y=75/100=7.5

the most economical dimensions of the garden would be:
 10 ft x 7.5 ft
width=7.5 ft
length=10 ft
and the cost when the area is boundened would be: $240.

C(x)=(12x²+1200)/x
C(10)=$240