# How much fencing is needed to define two adjacent rectangular playgrounds of the same width and total area 15,000 square feet?

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### 4 Answers

If I understood your problem correctly, we have two adjacent rectangles with the same width but different length. Since the problem is stating that we are defining the playground, we will have to assume that they want to place a piece of fence between the two restangles.

Let x be the width, then the length is 15,000/x. So the expression for finding the needed fence is

`P(x)=3x+2*15000/x => P(x)=3x+30000/x`

Let's find the 1st and 2nd derivatives of this function, to determine if we have min or max.

`P'(x)=3-30000/x^2`

`P''(x)=-30000*(-2)/x^3=-60000/x^3`

To find the max or min, we solve for P'(x)=0

`P'(x)=0 => 3-30000/x^2=0 => 3=30000/x^2=>`

`x^2=30000/3=10000 => x=+-100`

Since we are working with real dimensions we do not need to worry about negative value, we are only interested with x=100

At x=100, P"(x) is negative so the value is a min. so at the point 100, we can find the minimum possible amount of fence needed.

`P(100)=3*100+30000/100=300+300=600`

Hence the minimum needed is 600 ft.

It is not 3x+2, it is 3x+2*15000/x, three width (3x) and two of the length (2*15000/x)

The given information about the two play grounds are not sufficient. So this solution may be suitable to a particular situation. Even with the given given conditions, the dimensions of the rectangular play grounds could vary.So we assume the play grounds are are adjacent along the width w = x. And their lengths may be l1 and l2. So the entire Area = (l1+l2)x = A.

Therefore (l1+l2) = A/x.

The fencing is for the primeter and to define their separating common width = 2(l1+l2+x) + x = 2(l1+l2) + 3x.

P(x) = 2A/x +3x.

We now find the minimum value of P(x), which could be at x= c for which p'(c) = 0 and p''(c) < 0.

Differentiating , we get P'(x) = -2A/x^2 +3. Equating p'(x) = 0, we get:

-2A/x^2 + 0 , or 3x^2 = 2A, Or x = c = (2A/3)^(1/2).

We calculate p"(c):

P"(x) = (-2A/x^2+3)' = (-2/3)(-2/x^3) = 4/3x^3 which is positive for c = (2A/3.)^(1/2).

Therefore p (c) =2(l1+l2) +3c = 2 (A/c) +3c = 2A/(2A/3)^(1/2)+3(2A/3)^(1/2) = 2 (6A)^(1/2) is the minimum fencing required.

Put A = 15000 sq ft.

Then P(c) = 2(6*15000)^(1/2) = 2*300 = 600 feet minimum fencing is required, when width x= c= (2A/3)^(1/2) = (2*15000/3)^(1/2) = 100 feet. l1+l2 = A/c = 15000/100 = 150 feet.

For any other choice of measurement with the same area and a different common width we require a fencing much higher than 600 feet.

Rcmath,

Thank you for your help! I understand everything except for the 3x + 2? Where did this come from?

Thanks again!