# Calculus is killing me!

• Apr 8th 2008, 07:23 PM
Raina
Calculus 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, 07:27 PM
Mathstud28
OK here you go
Quote:

Originally Posted by Raina
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?

You have this $P_{field}=400=x+2y$ only one y becuase one side is missing...and $Area_{field}=xy$...using substitution you have tha $A=y\cdot(400-2y)$...diffentiating we get $A'=400-4y$... $y=50$...now we know this is a max since $A''<0,for every y since it is,A''-4$...so we know that $y=50$ is a max...so now put that back into your permiter equation and get x
• Apr 13th 2008, 09:14 PM
Raina
i noticed sth
when u slove 400-4y=0
u get 100
NOT 50...

any way thanks for ur time and help
• Apr 14th 2008, 12:11 AM
Moo
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 :)