What is the maximum area possible? Explain.?

Question

The city plans to enclose a rectangular area along the waterfront of Lake Eerie and create a park and swimming area. The budget calls for a purchase of 3000 ft of fencing. There is no fencing along the lake.

A(x)=x(3000-2x)

A.) What is the maximum area that can be enclosed? Explain.

Answer 1

Multiply through the parentheses:

A(x) = 3000x - 2x²

Now, take the first derivative of this equation,set it equal to zero, and solve for x:

A'(x) = 3000 - 4x = 0
4x = 3000
x = 750

So, the area of the space would be 750 x 1500 = 1,125,000 sq ft.

You integrate and set = 0 because you are essentially graphing the possible areas based on the length of one of the sides. When the tangent to the graph goes to zero, then the curve is at its maximum location.

Answer 2

your equation is all you need to calculate the maximum area: this equation is a function that tells you the area of the rectangle given a length of x and a width of 3000 - 2x.

remember that area = length x width (A = L*W)
you are not given either L nor W, so you pick one (lets say length) and assign it as x. so now we have:

A = x * W

now we still don't have enough information to solve the problem because we still have two unknowns, x and W. so let's express W in terms of x so we only have one uknown:

if x is the length, that means we used x feet out of 3000, so instead of having 3000 feet to make the width, we have 3000 - x. Also remember that a rectangle has FOUR sides, TWO of which are the length, and two are the width. So, we have to use 2x to make the length (the top and the bottom), meaning that we have 3000 - 2x left for the width. so L is x feet, and W is 3000 - 2x feet:

A = L * W
A = 2x * (3000 - 2x)
A = 6000x - 4x^2

this is a quadratic equation that will give you the area of the rectangle given a choice of length x. so let's say you pick the length to be 10:

A = 6000*(10) - 4(10)^2
A = 59600 sq ft

you can pick different values and A will get larger or smaller depending on what you choose to be L. if you use a graphing calcluator (or use algebra to find the midpoint of the parabola graph) you will find that the largest point happens when length = width, which is at L = W = 750 ft:

A = 6000*(1500) - 4(1500)^2
A = 2,250,000 sq ft

which happens to be a square! so the largest area enclosed by a rectangle is a square

hope that was helpful
-SelArom

Answer 3

Assuming that your boss would not choose an quite great area surrounded via an quite small fence there is not any think approximately reducing your roll up. What in the worldwide is all this width, length,area stuff? Is the area in which you're to place your fence of enormous quantity and flat? if so make a around fence. What has A(w) = w(220-w) to do with something? Ah, I see, it relatively is finding the area of a rectangle of width w and outer edge 440 and it relatively is maximised while w=a hundred and ten - yet this is basically for rectangles, circles are greater suitable. the area of a circle of circumference =440 is (440/pi)^2/4*pi = 220^2/pi .

Answer 4

You need to find the peak of the parabolic curve for the area based on the given perimeter
In this case it would be 750 ft x 1500 ft
750 x 1500 would be 1,125,000 sq ft
749 x 1502 would be 1,124,998 sq ft
751 x 1498 would be 1,124,998 sq ft

Answer 5

Set the first derivative of x to zero:

dA/dx = 3000 - 2x^2, solve for x:

x = 750, subst this into A(x) to get max Area of
max A = (750)(1500)

Answer 6

each small leg (side of rectangle) = x
long side = (total fence - 2 legs) (3000 - 2x)

area = 3000x - 2x^2 = max
1st deriv = 3000 - 4x (max / min happens @ vertex) = 0
3000 = 4x
x = 750(small side) .... so long side = (3000 - 2(750)) = 1500

750 * 1500 = 1,125,000 sq '

Answer 7

Same as before,

A(x) = x(3000 - 2x)

dA/dx = 3000 - 4x

Let dA/dx = 0

3000 - 4x = 0
x = 750

A(x) = 750 * 1500 = 1125000