I need an immediete help in this math problem?

Question

A gardener has 200 meters of fencing to enclose two adjacent rectangular plots what dimensions will produced a maximum enclosed area??



Could you please help me with this problem and please put your solution or an explanation on how did you get the answer

Thanks ^_^

Answer 1

First start by writing equations and variables:

you want to maximize area (A)
max A = W (width) * L (length)
2L + 3W = 200
A is maximized when the first derivative is equal to 0

You need to start by getting max A = W*L into a single variable. To do so, solve for L in terms of W in the 2L +3W =200 equation. You will get: L = 100 - 1.5W.

Substitute that into W*L = max A and you will get:
W (100 - 1.5W) = A

Expand and you get: 100W - 1.5W^2 = A
Move A to the other side, and take the derivative and set it equal to 0.

100W - 1.5W^2 - A = 0

Derivative: 100 - 3W = 0

W = 100 / 3 (or 33.333)

Put this into 2L + 3W = 200

2L + 3 (100/3) = 200

2L = 100

L = 50

Therfore, W= 33.333 , L= 50, and A = 1666.667

Hope this helps!

Answer 2

Two rectangular boxes side-by-side provides a top and a bottom width (w) and three side heights (h). The area of the combined boxes is....h*w

We know that 3*h + 2*w = 200, and
h * w = A

We want Amax, which will require a derivative, but we can't do a derivative with two variables, so let's simplify 3*h + 2*w = 200 for w and substitute it into h * w = A:

3*h + 2*w = 200
2*w = 200 - 3*h
w = 100 - 3*h/2

Substitute into l * w = A:
h * (100 - 3*h/2) = A
100*h -3*(h^2)/2 = A(h)

The maximum of A(h) can be found where the derivative of A(h) with respect to h, or d[A(h)]/dh = 0. And

d[A(h)]/dh = 100 - 3*h (assuming you can do derivatives)

So the value for h where 100 - 3*h = 0 will provide the maximum A.

100 - 3*h = 0
3*h = 100
h = 33.33333

Using 3*h + 2*w = 200, we can determine w and find the area.

3*33.333 + 2*w = 200
100 + 2*w = 200
2*w = 100
w = 50

Amax = h * w (for our derived values)
Amax = 33.333 * 50
Amax = 1666.67 meters squared

Answer 3

Is this for calculus? If it is, it is to illustrate a basic concept.

The perimeter is constant at 200m. If the rectangular sides are a and b, then

2(a+b) = 200
a+b = 100 (a constant)

the area would be A = a * b
A = a * (100 - a) substituting for b.
A = 100a - a^2
now differentiate

dA/da = 100 - 2a and that will equal zero at turning points (draw the graph if you don't understand what i'm talking about).

100-2a = 0 gives a=50.

d2A/da2 = -2 which indicates that a=50 is a maximum.

Answer 4

lets x the length and y the width

you have 2x+2y =200 or x+y =100 so y=100-x

the area is xy =A or x(100-x) =A

A = 100 x -x^2

you derive this dA/dx= 100-2x

the maximum when the derivative =0

x=50

Area is maximum for the quadrat 4 sides equal to 50 m

Answer 5

Depends on how big the lots are......If you can't enclose the entire perimeter.....then a circular fence would enclose the greatest area.

circumference= 2* Pi * R
200=2* 3.14* R
R=32

Area= Pi*r squared
3.14* 32*32= 3215 square meters

Answer 6

only square ( a special type of rectangle) will give the maximum area.
since there are 2 plots, area of one = 200/2 = 100 m^2

side of one square = sqrt (100)

dimension of each plot = 10meters.

Answer 7

49x49x49x49 and 1x1x1x1

Maximize one of the rectangles and make both into a square.