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
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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.
-x^+40x can also be simplified using completing the square method
-1(x^2-40x+400-400)=0 (add the suare of -40x divided by 2)
Therefore, the dimension are 20*20.
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