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?
Ok, thanks so much for your explanations.
So, is the max area 8,000,000?
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.
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.
can you help me with this homework please
What don't you understand?
No, you don't have to consider it as a triangle. Since the given problem states that the plot borders on a river and the farmer does not fence the side along the river, we can consider the plot as a rectangle with one side open to the river.
To find the largest area that can be enclosed, we need to determine the dimensions of the rectangle that will maximize the area. Let's call the length of the rectangle L and the width of the rectangle W.
Now, let's analyze the given information. The farmer has 8000 meters of fencing, which means the total length of the fencing will be the sum of all four sides of the rectangle: L + W + L + W = 8000.
Simplifying the equation, we get 2L + 2W = 8000. Dividing both sides by 2 gives us L + W = 4000.
Since the length L is along the river and does not require fencing, we can rewrite the equation as L + W + W = 4000, which simplifies to L + 2W = 4000.
Now, we need to express the area of the rectangle in terms of L and W. The area of a rectangle is calculated by multiplying its length by its width, so the area A = L * W.
Since we want to find the largest area, we can express the area A in terms of a single variable, taking into account the relationship we found between L and W earlier. Substituting L = 4000 - 2W into the area equation, we get A = (4000 - 2W) * W.
To maximize the area, we need to find the value of W that maximizes the expression A = (4000 - 2W) * W. One way to do this is by making a table to test different values of W and calculate the corresponding areas.
However, to simplify calculations, we can use calculus. Taking the derivative of A with respect to W and setting it equal to 0 will give us the critical points where the area is maximized. We can then solve for W to find the width that yields the largest area.
Differentiating A = (4000 - 2W) * W, we get dA/dW = 4000 - 4W. Setting this equal to 0 gives us 4000 - 4W = 0. Solving for W, we find W = 1000.
Now that we know W = 1000, we can substitute this value back into the equation L + 2W = 4000 to find L. L + 2(1000) = 4000, which yields L = 2000.
Therefore, the dimensions of the rectangle that will yield the largest area are L = 2000 meters and W = 1000 meters. Substituting these values into the area equation A = L * W, we find the largest enclosed area is 2,000,000 square meters.
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.