Three sides of a rectangular playing field are to be fenced with 400 m of fencing. Find the dimensions so that the area of the field will be a maximum.

Can any one solve this question?

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- Apr 8th 2008, 06:23 PMRainaCalculus is killing me!
Three sides of a rectangular playing field are to be fenced with 400 m of fencing. Find the dimensions so that the area of the field will be a maximum.

Can any one solve this question? - Apr 8th 2008, 06:27 PMMathstud28OK here you go
You have this $\displaystyle P_{field}=400=x+2y$ only one y becuase one side is missing...and $\displaystyle Area_{field}=xy$...using substitution you have tha $\displaystyle A=y\cdot(400-2y)$...diffentiating we get $\displaystyle A'=400-4y$...$\displaystyle y=50$...now we know this is a max since $\displaystyle A''<0,for every y since it is,A''-4$...so we know that $\displaystyle y=50$ is a max...so now put that back into your permiter equation and get x

- Apr 13th 2008, 08:14 PMRaina
i noticed sth

when u slove 400-4y=0

u get 100

NOT 50...

any way thanks for ur time and help - Apr 13th 2008, 11:11 PMMoo
Hello,

Let x and y be the sides (don't know how to name it :D) such as the perimeter P is :

P=x+2y=400 (one side is missing) -> x=400-2y

Plus, we want the area be at its maximum. The area A is :

A=xy

-> A=(400-2y)y=-2y²+400y

If you derivate : A'=-4y+400 and if y<100, A' will be positive, if y>100, A' will be negative. So it firstly increases and then decreases.

y=100 is a maximum, and so x=400-200=200 :)