Differentiation problem?

Question

"A fence delimits a patch from three sides: The front and two sides (in the back of the patch there's a wall).
The overall length of the fence is 40 feet.

What has to be the length of the front, so that the area of the patch will be maximal?"

How do you solve the question above?

Answer 1

let the length of the front be x
width will be (40-x)/2
area=x(40-x)/2
20x-x^2/2
for A max dA/dx=0
20-x=0
x=20
theanswer is 20'

Answer 2

Area = x.y
where x is the front side and y is the sides
And x+2y = 40 (length of fence)
So Area = x.1/2(40- x) where y = 1/2(40-x)
= 20x - 1/2x^2

Now the area is a maximum if dA/dx = 0 ( differentiation)
dA/dx = 20 - x
But this is equal to 0 when the area is maximum
Then 20 - x = 0
The length of the front x = 20 feet.

:-)

Answer 3

Let front be 2x to avoid fraction
then the 2 sides are 20-x
we have to maximize
x(20-x)
let y = x(20-x) = 20x - x^2
dy/dx = 20-2x = 0
d^y/dx^2 = -2x -ve

so maximum when front = 20
This can be solved without differentiation also
20x-x^2 = 100-(100 -20x +x^2) add 100 to convert to square
= 100-(x-10)^2 maximum when x = 10 then (x-10)^2 =0

Answer 4

This link will give you a full tutorial, and shows all the work for an area maximization using derivatives.

Regards,

Mysstere

Answer 5

same length as the wall