My kid has a math problem and he made it mine, help?

Question

word problem. . . A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 200 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?

It is an algebra question.

Answer 1

Say the side there's two of is the length L and the other side is the width W. The total perimeter is:

2L + W = 200

and the area is

A = LW

Substituting W = 200-2L, we have:

A = L(200 - 2L)

A = -2L² + 200L

This is a quadratic equation, and the maximum value of A is achieved at the vertex. The value of L at the vertex is:

L = -b / 2a

where a = -2 and b = 200.

That means L = -200 / (2*-2) = 50

From that we have:

W = 200 - 2L = 100.

Thus the maximum is in a 100 x 50 fence with area 5000 ft².

Please note that the maximum area is NOT a square because we are NOT surrounding the entire area with fence. After all, all those people saying it's a square keep getting 4444 ft², which is smaller than 5000 ft².

This is the correct solution, I've shown how to do it using algebra your child should know, and I was the first to do so. Hope this was helpful (at least moreso than the incorrect or unexplained answers).

Answer 2

draw a diagram

let x be the widths and y be the length

because there is one side that is not enclosed by the fence, there are only 3 sides that the fence form.

200 = x + x + y
200 = 2x + y

area = xy

solve for y
200 = 2x + y
y = 200 - 2x

area = x (200 - 2x)
area = 200x - 2x^2

the maximun area is the y value of the vertex. But first, find the x value of the vertex

-b/(2a)
-200 / (2*-2)
-200/-4
50

area = 200(50) - 2(50)^2
area = 10000 - 5000
area = 5000 ft^2

the maxinum area is 5000 ft^2

Answer 3

Let the width of the rectangle be x ft, so the length of it will be [200 - 2x] ft. The area [A] of this rectangle is x[200 - x] sq ft, so

A = x[200 - x]

= 200x - x^2

dA/dx = 200 - 4x, and at anymax or min point, dA/dx = 0, and

d2A/dx^ = -4, hence a max value.

200 - 4x = 0

200 = 4x

x = 50, ft, and the area = 50 x 100 = 5000 sq ft.

Answer 4

The problem can't be done with ordinary algebra. But basic calculus shows that the maximum area is achieved when the side parallel to the street is twice as long as the other sides. That would be 100 by 50.

Answer 5

The rectangle that encloses the most area using the smallest perimeter is a square. So, if you don't need to fence one side, you have to divide your 200 feet of fencing among three sides of a square. This means the side of the square will be 200/3 = 66.7 feet long.

The area of a square is s^2, where s is the side length.

A = s^2
A = (66.7 ft)^2
A = 4444 ft^2

So, your fence (and the street on one side) encloses 4444 square feet.

Answer 6

The perimeter of this three sided fence will be one length plus two sides and that has to add up to 200. The most coverage would be if the sides of the rectangle are equal. If you have three equal sides that add up to 200 then you get the equation
3L=200 so L= 200/3 which equals 66 2/3 feet for the length and for each of the two sides.

Answer 7

Three sides sum to 200 feet. A square is the largest area/perimeter. Do it.

1) Consider making two end sides shorter and the middle side longer. The limit is zero area.

2) Consider making two end sides longer and the middle side shorter. The limit is zero area.

3) The average is the maximum.

Answer 8

Would it be 50ft per side? I don't know how to put it in proper terms and words. Hope that works? Or at least helps.

Actually I take that back- The whole "not fence the side along street" part doesn't click for me. Sorry I don't know how to solve that problem without knowing how long, in ft, the street is.

Answer 9

50x100x50 would be the dimensions and that adds to 200 feet and 100 times 50 =5000 so its 5000 square feet

Answer 10

for minimum P
L = 2w
Perimeter = 200 = 2w + L = 2w + 2w
4w =200
w = 50
L = 100