You have two formulas to use:
A = lw -->area = length times width
P = 2l + 2w -->perimeter
You know the perimeter, which is 200m of fencing.
Let length = x
Now let's solve the perimeter equation for w, using what we know:
200 = 2x + 2w
200 - 2x = 2w
(200 - 2x)/2 = w
2(100 - x)/2 = w
100 - x = w
Therefore:
A = x(100 - x) = 100x - x^2
-x^2 + 100x is the area of the rectangle.
To place the quadratic equation in vertex form, you will need to complete the square:
A = -x^2 + 100x
A = -(x^2 - 100x)
A = -(x^2 - 100x + 2500 - 2500) -->to complete the square, take 1/2 times the middle coefficient and square it (you'll need to add and subtract it as well)
A = -(x - 50)^2 + 2500 -->vertex form
If x = 50, then 100 - x = 50.
To maximize the area, take (50)(50) = 2500 square meters.
I hope this will help with other problems of this type.
Ms. Brown has 200m of fencing with which she intends to construct a rectangular enclosure along a river where no fencing is needed. She plans to divide the enclosure into two parts.
What should be the dimensions of the enclosure if its area is to be a maximum?
Show a complete algebraic solution.
thank you
blah blah blah
IM NOTHING
To find the dimensions of the enclosure that will maximize its area, we can use the given formulas:
A = lw (area = length times width)
P = 2l + 2w (perimeter)
We are given the perimeter, which is 200m of fencing. Let's assign a variable for the length, so let's say length = x.
We can now solve the perimeter equation for width using what we know:
200 = 2x + 2w
200 - 2x = 2w
(200 - 2x)/2 = w
2(100 - x)/2 = w
100 - x = w
Therefore, the width of the enclosure is equal to 100 - x.
Now, let's express the area of the enclosure using the dimensions:
A = x(100 - x) = 100x - x^2
The area of the enclosure, A, is equal to -x^2 + 100x.
To find the maximum area, let's put the quadratic equation in vertex form by completing the square:
A = -x^2 + 100x
A = -(x^2 - 100x)
A = -(x^2 - 100x + 2500 - 2500) -- to complete the square, take 1/2 times the middle coefficient and square it (you'll need to add and subtract it as well)
A = -(x - 50)^2 + 2500 -- vertex form
If x = 50, then 100 - x = 50.
To maximize the area, we take (50)(50) = 2500 square meters.
So, the dimensions of the enclosure that will maximize its area are length = 50m and width = 50m.
I hope this explanation helps with similar problems in the future.