Answer :
The minimum perimeter of the rectangle is 45.25m
How to find the minimum perimeter
The are of a rectangle is solved by Length * width
therefore
128 m² = L * w
L = 128/w
perimeter of a rectangle
P = 2L + 2w
P = 2(128/w) + 2w
P = 256/w + 2w
minimum perimeter = dP/dw
dP/dw = -256/w² + 2 = 0
256/w² = 2
256 = 2w²
w² = 128
w = 11.31 m
P = 256/w + 2w
substitute the value of w = 11.31 m
P = 256/11.31 + 2 * 11.31
P = 22.63 + 22.62
P = 45.25 m
the minimum perimeter is 45.25 m
Learn more about minimum perimeter at:
https://brainly.com/question/17390576
#SPJ1
Answer:
- perimeter: 32 m
- dimensions: 16 m parallel to the river; 8 m wide
Step-by-step explanation:
You want the length of the minimum perimeter fence to enclose 3 sides of a rectangular area of 128 m² with one side of the enclosure provided by a river. You also want the enclosure dimensions.
Perimeter
If x represents the dimension of the enclosure parallel to the river, then the other dimension of the enclosure is found from ...
A = LW
128 = x·W
W = 128/x
There will be two sides of this length, so the perimeter is ...
P = L +2W = x +2(128/x)
P = x + 256/x
Minimum
This has an extreme value (minimum) where its derivative is zero:
dP/dx = 0 = 1 -256/x²
Solving for x gives ...
x² = 256
x = √256 = 16
and the perimeter is ...
P = 16 +256/16 = 32
The other dimension is 128/16 = 8.
The minimum perimeter is 32 meters; the enclosure is 16 by 8 meters.
__
Additional comment
This can be solved without using derivatives by working the reverse problem: the largest area for a given perimeter.
Using the same definition of x, the area in terms of perimeter is ...
A = x · (P -x)/2
This is a quadratic equation with a maximum halfway between its zeros at x=0 and x=P. So, the value of x for maximum area is x = P/2. That area is ...
A = (P/2)(P -P/2)/2 = (P/2)(P/4) = P²/8
Or, the minimum perimeter for a given area is ...
P = √(8A)
In this problem, that is ...
P = √(8·128) = √1024 = 32
The dimensions of the enclosure are (P/2)×(P/4) = 16×8.
