Maximum rectangular area...?

Question

Jose has 100 feet of fence. What is the maximum rectangular area that he can enclose with this much fence?

Forgot how to do these kind of problems too. =p

Answer 1

the best bet is always a square. if he has 100 ft, find a quarter of that, which is 25 on each side.

Answer 2

Total length = 100

let the sides of the rectangular be x & y.

Hence, the area of the rectangular is A = xy

We know that:

2x+2y = 100 (i.e. the perimeter)

re-write:

y = 50 - x

substitute y into A = xy:

A = x(50-x) = 50x - x^2

for maximum or minimum, take derivative of A with respective to x:

dA/dx = d (50x - x^2) /dx

= 50 - 2x

set dA/dx = 0:

50-2x - 0

solve for x:

x = 25

However, x = 25 could be maximum or minimum value.

You need to check whether it is max or min by taking 2nd derivative of A:

d^2 A / dx^2 = -2 < 0  gives dA/dx to be the maximum value

x = 25 is OK

y = 50 - 25 = 25

It is a square :)

A = 25 x 25 = 625

Answer 3

2L + 2W = 100 ft

L + W = 50

W = 50 - L

area = LW

= L(50-L)

= 50L - L²

Maximum area when first derivative = 0.

area' = 50-2L = 0

L = 25

W = 50-L = 25

Maximum area = 25×25 = 625 ft²

Answer 4

1) Setup the perimeter equation.

2L + 2W = 100

2) Solve for either L or W.

L = (100 - 2W)/2

3) Setup the area equation.

LW = A

4) Plug in L in terms of W (step 2), into the area equation.

W(100 - 2W)/2 = A

(100W - 2W^2)/2 = A

-W^2 + 50W = A

5) Find the maximum (vertex) of the parabolic-like function. The vertex can be given by -b / 2a.

-50 / 2(-1)

25

6) Plug in you maximum (vertex) back into the function to find your area.

-(25)^2 + 50(25) = A

-625 + 1250 = A

625 = A

The maximum area that the rectangle can have is 625 feet^2.

Answer 5

It would be a square, each side 25 ft.

Area = 25² or 625 ft²