A rectangular enclosure is to be constructed with 800 feet of fence in order to have four sections. What should be the dimensions in order to maximize the area enclosed? What is the maximum area?
Well, I bet it turns out to be a square but anyway:
A = b c where b is breadth and c is length
2 b + 2 c = 800
so
b + c = 400
c = (400 - b)
so
A = b * (400-b)
A = 400 b - b^2
b^2 - 400 b + A = 0
where is the vertex of this parabola (I assume you do not do calculus)
b^2 - 400 b = -A
b^2 - 400 b + 40,000 = -A + 40,000
(b-200)^2 = -A + 40,00
vertex (maximum) when b = 200
then c = 400-200 also = 200
just as we suspected
To find the dimensions that will maximize the enclosed area of a rectangular enclosure, we can use calculus.
Let's denote the length of the enclosure as L and the width as W.
The perimeter of the rectangular enclosure is given as 800 feet, so we have the equation:
2L + 2W = 800 (equation 1)
We want to maximize the area, and the area of a rectangle is given by:
A = L * W
To proceed, we need to eliminate one variable (either L or W) from the equation. We can solve equation 1 for L:
L = (800 - 2W) / 2
L = 400 - W (equation 2)
Now we can substitute this value of L into the equation for area:
A = (400 - W) * W
A = 400W - W^2
To maximize the area, we take the derivative of A with respect to W and set it equal to zero:
dA/dW = 400 - 2W = 0
Solving this equation:
400 - 2W = 0
2W = 400
W = 200
Substituting this value of W back into equation 2 to find L:
L = 400 - W
L = 400 - 200
L = 200
So, the dimensions that will maximize the enclosed area are a length of 200 feet and a width of 200 feet.
To find the maximum area, substitute these values into the equation for area:
A = L * W
A = 200 * 200
A = 40000 square feet
Therefore, the maximum area enclosed is 40,000 square feet.