No, make the river one of the side, fence the other three sides.
Area= LW
8000=2W+L
or L=8000-2W
Area= W(8000-2W)= 8000w-2W^2
You can find the max several ways, graphing is simple. IF you get stuck, repost.
A farmer with 8000 meters of fencing wants to enclose a rectangular plot that borders on a river. If the farmer does not fence the side along the river, what is the largest area that can be enclosed?
Does that mean I have to consider it a triangle?
6 answers
Thanks. I'm still confused though. I'm not sure how you got
Area= W(8000-2W)= 8000w-2W^2
What happened to the L
Do I need a system of equations?
Sorry, this has got me stumped.
Area= W(8000-2W)= 8000w-2W^2
What happened to the L
Do I need a system of equations?
Sorry, this has got me stumped.
It is a quadratic.
Let y= 8000x-2x^2
graph y vs X on your graphing calc, notice where the max is on x
Second method. The parabola goes up to a max then down. Find the intercepts for y=0, those will be symettrical to the parabolic axis, so look for where the midpoint of the intercepts are.
y=x(8000-2x)
intercepts x=0 , x=4000, so the max will be at x (or width 2000).
then solve for L (8000-2W).
Third method:
Calculus (in a few years you will master this, just watch now)
Area= 8000x-2x^2
d Area/dx=0= 8000-4x
solve for x, x=2000 at max.
Let y= 8000x-2x^2
graph y vs X on your graphing calc, notice where the max is on x
Second method. The parabola goes up to a max then down. Find the intercepts for y=0, those will be symettrical to the parabolic axis, so look for where the midpoint of the intercepts are.
y=x(8000-2x)
intercepts x=0 , x=4000, so the max will be at x (or width 2000).
then solve for L (8000-2W).
Third method:
Calculus (in a few years you will master this, just watch now)
Area= 8000x-2x^2
d Area/dx=0= 8000-4x
solve for x, x=2000 at max.
Ok, thanks so much for your explanations.
So, is the max area 8,000,000?
So, is the max area 8,000,000?
can you help me with this homework please
What don't you understand?
1 answer