fencing?

[PLAYNOISE]

ok this is the near last step of the problem i was working on earlier but it doesnt have much to do with it and im confused on what to do so here it goes

Suppose the rectangular plot runs along a building, so that you need to fence only three sides. What dimensions will now yield a maximum area with 100 meteres of fencing?

please help! thanks so much!

 

[Gene]

It is L long and W wide.
The Area = LW
The fencing is L+2W=100
Solve for L and substitite in A = LW
You get A = -2W²+100W
The maximum is at x=-b/2a in ax²+bx+c

(I corrected w to W. W & w are two different variables. )

 

[PLAYNOISE]

ok so is length was x and width was y (opposed to L and W) would the formula be A= -2y^2 + 100y? And then do i need to solve that further? And then would the maximum be -100y / 2(-2y)? I'm still confused about the solving portions.thanks!

 

[Gene]

The equation is -b/2a for the maximum (or minimum) of ax2+bx+c. The variable is not part of it. Just the constants. -b/2a is derived from the quadratic equation and works for any quadratic.
In this one
a=2
b=100
c=0
So the maximum is at
W = -100/-4
You plug W into the original equation to find L.

Just a hint. It is better to use variable names that help to keep in mind what it means. That's why I chose L & W. When you get to the end of a long computation it is hard to know what x, y, and z were but L, W and H in a cube make it easy.