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

Asked on by laura003

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hala718's profile pic

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's profile pic

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.

 

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