Question

A promoter wants to enclose two identical seating areas on the beach for VIP seating for a surfing competition. One side of the seating areas is along the ocean and does not require any fencing. If the total area for the two seating areas must be 19,200 ft2, find the dimensions which will minimize the amount of fencing to be used. Ocean (no fencing)

Answer 1

Dimensions of each seating areas:

x = 120 ft

y = 80 ft

p(min) = 320 ft

Explanation:

A = 19200 ft²

Let´s call "x" the sid of the fence parallel to the ocean ( only one side will be fenced), then the perimeter of the whole area is:

p = x + 3*y ( y equal, two plus the splitting side )

A = x*y = 19200

y = 19200/ x

The perimeter as a function of x

p(x) = x + 3* ( 19200/x)

Taking derivatives on both sides of the equation

p´(x) = 1 - 57600/x²

p´(x) = 0 ⇒ 1 - 57600/x² = 0

x² - 57600 = 0

x = √ 57600

x = 240 ft

and

y = A/x

y = 19200/ 240

y = 80 ft

We look at the first derivative, by subtitution of x = 240 the first derivative p´(x) > 0 therefore we have a minimun for p(x) at the value x = 240

Then x/2 = 120 ft

and y = 80 ft

fencing material minimum = 80 + 3*80

F(min)= 320 ft