Respuesta :
So long as the perimeters are the same, rectangles and squares share the same area. For example, a square that is 2m by 2m across is 4m squared. A rectangle of 4m by 1m across is still 4m squared.
Therefore all we want to do here is see how big we can make our “square” perimeter using the creek. We have three sides to spread 580ft across, therefore if we divide this by 3, we get 193.3ft of fencing per side. If we then square this figure, we will then get the maximum possible area, which comes to 37,377ft squared. (That’s a huge garden).
Therefore all we want to do here is see how big we can make our “square” perimeter using the creek. We have three sides to spread 580ft across, therefore if we divide this by 3, we get 193.3ft of fencing per side. If we then square this figure, we will then get the maximum possible area, which comes to 37,377ft squared. (That’s a huge garden).
Answer:
Maximum possible area of the corral will be 42050 square feet.
Step-by-step explanation:
Monique wants to enclose a rectangular area, using creek on one side and other three sides by the fence.
Let the length of one side of the rectangular area is 'x' feet and other side is 'y'.
Length of the fencing has been given as 580 feet.
Therefore, (2x + y) = 580
2x + y = 580
y = (580 - 2x)-------(1)
Area of the rectangular area = length × width
A = xy
Now we replace the value of y in the area
A = x(580 - 2x)
= 580x - 2x²
For the maximum area of the corral, we will find the derivative of the area A with respect to x and equate it to zero.
[tex]\frac{dA}{dx}=\frac{d}{dx}(580x-2x^{2})[/tex] = 0
580 - 4x = 0
4x = 580
[tex]x=\frac{580}{4}[/tex]
x = 145 feet
From equation (1)
y = (580 - 2×145)
= 580 - 290
= 290 feet
Maximum area covered of the corral = xy
= 145×290
= 42050 square feet
Therefore, maximum possible area of the corral will be 42050 square feet.
