A rectengular area is to be enclosed on three sides by fencing, an existing hedge forming the fourth side. Find the greatest area of the rectangle and its dimensions if 140mof fencing is available.
let the length parallel to the hedge be y
let the width be x
so y + 2x = 140
y = 140-2x
area = xy
= x(140-2x) = -2x^2 + 140x
if you know Calculus:
-4x + 140 = 0
x = 35
then y = 140 - 70 = 70
max area = 35(70) = 2450
If no Calculus:
complete the square to find the vertex
A = -2x^2 + 140x
= -2(x^2 - 70x)
= -2(x^2 - 70x + 1225 - 1225)
= -2( (x-35)^2 - 1225)
= -2(x-35)^2 + 2450
so the vertex is (35,2450)
giving us a max of 2450 when x = 35
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To find the greatest area of the rectangle, we need to determine the dimensions of the rectangle that maximize its area.
Let's assume the length of the rectangle is x meters. In this case, the width of the rectangle should be (140 - 2x) meters to account for the three sides that need fencing.
The formula for the area of a rectangle is A = length * width. Therefore, the area of the rectangle in terms of x is A = x * (140 - 2x).
To maximize the area, we can find the value of x that corresponds to the maximum area. We can do this by taking the derivative of the area function with respect to x and setting it equal to zero.
dA/dx = 140 - 4x = 0
Solving this equation, we find that x = 35.
Now we can substitute this value back into the area equation to find the maximum area:
A = 35 * (140 - 2*35)
A = 35 * 70
A = 2450 square meters
Therefore, the greatest area of the rectangle is 2450 square meters, and its dimensions are 35 meters in length and 70 meters in width.