If 85 meter wire is used to cover a rectangular plot, then maximum possible area of the plot covered

let x as length y as breadth
by given 2(x+y)=85 ,
then for max possible value x=y
2 * 2x = 85
x=21.25
area = 21.25 * 21.25 = 451.5625

2x+2y=85
x+y=42.5
maxmaximum means x=y
2x=42.5
x=21.25.
maximum area=a*a=451.5625.

If sides of plot are x & y respectively
Perimeter of plot=85 =2(x+y), x+y=42.5
Maximum area is possible with maximum length of side,If the plot size is square,maximum side length=42.5/2=21.25
So maximum area=21.25^2= 451.5625 m^2

2L+2W=85;
L+W=42.5;
A=L*W;
A=(42.5-W)*w
=42.5W-W^2;
=-W^2+42.5W;------>(1)
Now it is in the format of ax^2+bx+c;
a=-1;b=42.5
So, now we have to fond the vertex.W=-b/2a
W=21.25;
Now sub in eqn. 1
So, A=-(21.25^2)+42.5(21.25)
A=451.5625

let the length be L and the width be W. I have 85 meter wire is used to cover a rectangular plot, so the perimter equation is:

2L + 2W = 85

Dividing by 2 to make things simpler, I get:

L + W = 85/2


A = L × W

I can substitute for either one of these variables by solving the perimeter equation:

L + W = 85/2
L = 85/2 – W (solving for L)

Then:
A = (85/2 – W) × W (substituting into the area equation)
= (85/2)W – W^ 2
There are a couple different ways of finding the vertex. I\'ll take the easy way. The equation of the quadratic, in y = ax^2 + bx + c format, is:

A = –W^2 + (85/2)W

The vertex of a parabola is the point (h, k), where h = –b/2a . In this case:

h = –(85/2)/(2×(–1)) = 85/4

To find the \"k\" part of the vertex, all I do is plug 85/4 in for W:


k = –(85/4)2 + (85/2)(85/4) = 451.625

My points from this equation are (W, A) — that is, I plug in a width and figure out the area — so the \"h\" is the maximizing width and the \"k\" is the maximum area. So the answer is:

The largest possible area is 451.5625 square meters

According to the given Conditions
Perimeter = 85 m
2(a+b) = 85

Let us assume length a =x m
b = 42.5 - x

a^2 + b^2 = 1/2[(a+b)^2 + (a-b)^2]
= 1/2[(42.5)^2 + (2x-42.5)^2]
= 2x^2 -85x + 1806.25
Therefore, x = 21.25 m

the length a = 21.25 m
the breadth b = 42.5-a = 42.5-21.25 = 21.25 m

So, The area of a rectangle is a.b =(21.25)(21.25)
= 451.5625 sq.m
the answer is 2.

Soln:
given that 85 meter wire is used to cover a rectangular plot ,this means that perimeter of rectangular plot is give.
we know that,perimeter of rectangle is 4l.
therefore, 4l=85
or, l=85/4 =21.25 meter
now,maximum possible area of rectangular plot is equal to l*l=(21.25)*(21.25)=451.5625 sq. meter.
Thus, our answer is 451.5625 which is option (2)

Given Length of the wire = 85 m
Perimeter = 2(a+b) = 85

Let us assume length l =x m
b = 42.5 - x

a^2 + b^2 = [(a+b)^2 + (a-b)^2]/2
= [(42.5)^2 + (x-(42.5-x))^2]/2
= 2x^2 -85x + 1806.25
Then, x = 21.25 m

l = 21.25 m
b = 42.5-l = 42.5-21.25 = 21.25 m

So, The area is l*b =(21.25)*(21.25)
= 451.5625 m^2
So,answer is 451.5625 m^2.

85 meter wire is used to cover a rectangular plot
so, 2*(x+y)= 85
=>x+y = 42.5
=>x = 42.5-y

so,the area is A=x*y
=>A=(42.5-y)*y
=>42.5*y-y^2
now differentiating both sides w.r.t y,we get
dA/dy= 42.5-2*y

now, for maximum dA/dy=0
so, 42.5- 2*y = 0
=>y=21.25
so,x=42.5-y=42.5-21.25
=>x=21.25
so,The maximum area is A=x*y
=>A=21.25*21.25= 451.5625
so,The Correct answer is 2
:)

since the wire can cover up the rectangle its circumference 2*(l+b)=85 => l+b=42.5
area is highest if l=b i.e 21.25*21.25=451.5625

Area will be max when the given rectangle is a square i.e. all sides are equal.
Then area = (85/4)^2= 7225/16 = 451.5625 sq mtrs

x:breadth,y:length
2*{x+y}=85
x=42.5-y
Now,A=42.5y-y^2
y=-42.5/2(-1)
y=21.25
Hence,A=42.5*21.25-21.25^2
A=451.5625

the length and breadth of the rectangle varies between 43 and 42.Between (43/2) and (42/2) the max product i.e l*b obtained in between those values is 451.5625

we wil get maximum area only when length=bredth
so length=bredth=22.25m and area is square of it

Let l=b
Perimeter of rectangle=85
=2*(l+l)=85
=4l=85
=l=85/4
=21.25
thus,area of rectangle=21.25*21.25
=451.56

perimeter of the rectangle plot = 2(l+b) = 85
=> (l+b) = 85/2 = 42.5
The maximum possibility of l & b to get lb is greatest l = b = 42.5/2 = 21.25
Ans so, lb = maximum area = 21.21 x 21.25 = 451.5625

Area is maximum if it is a square.

Perimeter of square = 85m = 4a
=> side of square = 21.25
=> Area = a^2 = 451.5625 sq.m

2(l+b)=85;
l+b=42.5;
2l=42.5;
l=21.25;
l*b=l*l;
21.25*21.25;
=451.5625

If the perimeter of the rectangle is 85 metre i.e
2(b+H)=85 then B+H = 42.5
SO the Max area of the rectangle i.e B X H
can be half i.e 21.25 X 21.25 or 451.5625

as length is 85 so diameter max is taken as 85/16 = 5.3125. so max area will be 85*5.3125=451.5625 i.e; option 2

2(L+B)=85
B=42.5-L
A=L*B
A=L(42.5-L)
differentiate this eqn w.r.t L and equate it to 0.
which will giv u value of L as 21.25
from L+B=42.5,FIND B
therefore A=L*B
A=21.25*21.25
A=451.5625

for a given length square covers maximum area. so 85/4=21.75 and area is 21.75*21.75=451.5625

2(l+b)=85;
b=(85/2)-l;
A=lb=l((85/2)-l);
for max differentiation of A with respect to l should be equal to zero. From this we get l=21.25 and b=21.25;

2(l+b)=85
l+b=42.5
l*b=?
maximum value of l*b is possible when l=b=21.25
l*b=451.5625

Let side length of the rectangle for maximum area be x;
then, area=x*((85/2)-x));
& for max
d(area)/dx=0;
so, x=85/4;
so,area=85/4*85/4
=7225/16
=451.5625 ans.
so, option 2 is a correct choice.

=85/4 =21.25

21.25*21.25=451.5625

Given perimeter of rectangle= 85 meters
i.e., l+b=85
To maximize the area,l=b
(proof can be given by first derivative rule of calculus)
Geometry says among all the rectangles, a square has the maximum area.
Therefore, l=85/4
area= (85/4)*(85/4)=451.5625

because area is length * breath
as given total perimeter of the rectangular box is 85,so maximum are is covered possible when all sides are equal so 85/4 is each side
so area is 451.5625

given 2l+2h=85m
find A=l*h
=> A=l*(85/2-l)
=> A=(85*l)/2-l^2
=> dA/dl=0
=> 85/2-2l=0
=> l=85/4
then h=85/4
area A= (85/4)*(85/4)
A= 451.5625 m^2

2(l+b)=85
l+b=42.5
to have max.. area let us consider
l=21.25+x
b=21.25-x
l*b=21.25^2-x^2
to have max area
let x=0
therefore....max area ==451.5625

as we know that,
for equal perimeter of all different rectangles, only a square maximum area, so
4p=85 --> p=21.25
nd area = p*p = 451.5625

max area=(85/4)=451.5625

Given the Perimeter of rectangular plot = 85
=> 2(l + b) = 85
=> (l+b) = 85/2 = 42.5

let l = x then b = 42.5-x

Area = (42.5 - x).x
differentiate it with respect to x and equate to 0
x = 21.25
so l = 21.25 and b = 21.25
so maximum area = 451.5625

2(x+y)=85;x+y=42.5;xy=21.5*21

l+b=42.5
area is max. when l=b=22.5
area=(22.5^2)=451.5625

Let the length be L and the width be W. I have 85 meter wire to cover a rectangular plot, so the perimter equation is:

2L + 2W = 85

Dividing by 2 to make things simpler, I get:

L + W = 42.5

We have to find the area; in particular, to find the largest area, given this perimeter. Let's look at the area equation:

A = L × W

I can substitute for either one of these variables by solving the perimeter equation:

L + W = 42.5
L = 42.5 – W (solving for L)

Then:

A = (42.5 – W) × W (substituting into the area equation)
= 42.5W – W^2
In other words, my area equation is a quadratic, and I'm supposed to find the maximum. So all I really need to do is find the vertex. Since the above area equation is a negative quadratic, then it graphs as an upside-down parabola, so the vertex is the maximum.

There are a couple different ways of finding the vertex. I'll take the easy way. The equation of the quadratic, in y = ax2 + bx + c format, is:

A = –W^2 + 42.5W

The vertex of a parabola is the point (h, k), where h = –b/2a . In this case:

h = –(42.5)/(2×(–1)) = 21.25

To find the "k" part of the vertex, all I do is plug 21.25 in for W:

k = –(21.25)^2 + 42.5(21.25) = 451.56 Answer

Let the length be L and the width be W. I have 85 meter wire to cover a rectangular plot, so the perimter equation is:

2L + 2W = 85

Dividing by 2 to make things simpler, I get:

L + W = 42.5

We have to find the area; in particular, to find the largest area, given this perimeter. Let's look at the area equation:

A = L × W

I can substitute for either one of these variables by solving the perimeter equation:

L + W = 42.5
L = 42.5 – W (solving for L)

Then:

A = (42.5 – W) × W (substituting into the area equation)
= 42.5W – W^2
In other words, my area equation is a quadratic, and I'm supposed to find the maximum. So all I really need to do is find the vertex. Since the above area equation is a negative quadratic, then it graphs as an upside-down parabola, so the vertex is the maximum.

There are a couple different ways of finding the vertex. I'll take the easy way. The equation of the quadratic, in y = ax2 + bx + c format, is:

A = –W^2 + 42.5W

The vertex of a parabola is the point (h, k), where h = –b/2a . In this case:

h = –(42.5)/(2×(–1)) = 21.25

To find the "k" part of the vertex, all I do is plug 21.25 in for W:

k = –(21.25)^2 + 42.5(21.25) = 451.56 Answer

Let the length be L and the width be W. I have 85 meter wire to cover a rectangular plot, so the perimter equation is:

2L + 2W = 85

Dividing by 2 to make things simpler, I get:

L + W = 42.5

We have to find the area; in particular, to find the largest area, given this perimeter. Let's look at the area equation:

A = L × W

I can substitute for either one of these variables by solving the perimeter equation:

L + W = 42.5
L = 42.5 – W (solving for L)

Then:

A = (42.5 – W) × W (substituting into the area equation)
= 42.5W – W^2
In other words, my area equation is a quadratic, and I'm supposed to find the maximum. So all I really need to do is find the vertex. Since the above area equation is a negative quadratic, then it graphs as an upside-down parabola, so the vertex is the maximum.

There are a couple different ways of finding the vertex. I'll take the easy way. The equation of the quadratic, in y = ax2 + bx + c format, is:

A = –W^2 + 42.5W

The vertex of a parabola is the point (h, k), where h = –b/2a . In this case:

h = –(42.5)/(2×(–1)) = 21.25

To find the "k" part of the vertex, all I do is plug 21.25 in for W:

k = –(21.25)^2 + 42.5(21.25) = 451.56 Answer