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 $\displaystyle P_{field}=400=x+2y$ 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
Hello,
Let x and y be the sides (don't know how to name it ) 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