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

3 answers

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.