Answer the area of a rectangle is Area=lengthxwidth
The length is 10metres and the width is 3 1/2metres or 3.5metres
If the length of the rectangle is 3m more than ×2 the width, the dimensions (length & width) that will give an area of 27m^2 =?
Firstly, the area of a rectangle = (L+W)×2
Since the length is 3 more than ×2 the width,
The length, L us therefore = (2×w)+3= 2w+3
Recall that the area is 27m^2
Now, substitute L for 2w+3
[(2w+3)+w]×2 = 27
(3w+3)×2 = 27
6w = 21
w = 21/6
w= 3 1/2m
If the width = 3 1/2, and area = (L+w)×2, we substitute "w" for 3 1/2 in the equation below:
(L+ 3 1/2)×2 = 27
2L + 7 = 27
2L = 20
L = 10m
Therefore, the length = 10m and the width
= 3 1/2m
3m and 9m
length of the rectangle = 2w+3
width of rectangle = w
area = L*W
if area = 27m²
using factorization w= 3 or -4.5
The answer is C.
The greatest area of rectangle is:
625 square units.
It is given that:
A rectangle of perimeter 100 units has the dimensions as:
50-w on the top and bottom.
and a w on its left side and right side.
i.e. we may say the length of the rectangle is:
and the width of the rectangle is:
Now, we need to find the greatest area of rectangle.
As the area of rectangle is:
A = w(50 - w)=50w-w^2
Now, to find the maximum area we differentiate the Area with respect to the width as:
Hence, to obtain the maximum area the width of the rectangle is: 25 units.
and that of the length of the rectangle is:
Hence, the dimensions of rectangle in order to obtain the maximum area is:
25 units by 25 units.
So, the area of rectangle is:
Hence, the greatest area of rectangle is:
625 square units.
to have the greatest area, try to make legnth and width the same
greatest is 625 square units
Since the acceleration is a constant negative, that means that when velocity, dA/dw=0, it is at an absolute maximum for A(w)...
dA/d2=0 only when 50=2w, w=25
So as the case with any rectangle, the perfect square will enclose the greatest area possible with respect to a given amount of material to enclose that area...
So the greatest area occurs when W=L=25 in this case:
Area maximum is thus: