# a rectangular yard is to be fenced in with 80 m of fence. what must the dimensions of the yard be in order to maximize the area of the field?this question was in my quadratic functions chapter in...

a rectangular yard is to be fenced in with 80 m of fence. what must the dimensions of the yard be in order to maximize the area of the field?

this question was in my quadratic functions chapter in the workbook

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Let the dimensions of the yard be x and y.

==> Given the perimeter is 80 m.

==> 2x + 2y = 80

==> x + y= 40

==> y= 40-x............(1)

Now we need to find the maximum area.

==> A= xy

But, from (1) , we know that y= 40-x.

`==gt A= x(40-x) = 40x - x^2 `

`==gt A(x)= -x^2 + 40x`

Now we will find the maximum value.

We notice that the sign of `x^2`  is negative, then, A(x) has a maximum area at A'(x) = 0

==> A'(x)= -2x + 40 = 0

==> -2x = -40

==> x = 20

==> y= 40-20 = 20

Then, the maximum area is when the dimensions are 20 X 20.

elekzy | Student, Grade 11 | (Level 1) Valedictorian

Posted on

-x^+40x can also be simplified using completing the square method

-1(x^2-40x)=0

-1(x^2-40x+400-400)=0   (add the suare of -40x divided by 2)

-1(x^2-40x+400)-1(-400)=0

-1(x^2-40x+400)+400=0

-1(x-20)^2+400=0

x=20

y=40-x

y=40-20=20

Therefore, the dimension are 20*20.