A regular 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

Answer: 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