Peter has 800 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum​ area?
200
To find the dimensions of the rectangle that maximize the enclosed area, we can use the concept of calculus. Let's assume that the length of the rectangle is x yards, and the width is y yards.
Given that the perimeter of a rectangle is equal to 2 times the length plus 2 times the width, we have:
2x + 2y = 800.
To solve for y in terms of x, we can rearrange the equation:
2y = 800 - 2x,
y = (800 - 2x)/2,
y = 400 - x.
Now, the area of a rectangle is given by the product of its length and width, so we have:
Area = x * y,
Area = x*(400 - x).
To maximize the area, we need to find the value of x that makes the derivative of the area with respect to x equal to zero. Let's differentiate the area function with respect to x:
d(Area)/dx = d(x*(400 - x))/dx,
d(Area)/dx = 400 - 2x.
Now, set the derivative equal to zero and solve for x:
400 - 2x = 0,
2x = 400,
x = 200.
So, the length of the rectangle that maximizes the area is 200 yards. Substituting this value back into the equation y = 400 - x, we can find the width:
y = 400 - 200,
y = 200.
Thus, the dimensions of the rectangle that maximize the enclosed area are a length of 200 yards and a width of 200 yards.
To find the maximum area, we can substitute the values of x and y into the area function:
Area = x * y,
Area = 200 * 200,
Area = 40,000 square yards.
Therefore, the maximum possible area is 40,000 square yards.
To find the dimensions of the rectangle that maximize the enclosed area, we can start by setting up an equation based on the given information. Let's represent the length of the rectangle as L and the width as W.
According to the problem, Peter has 800 yards of fencing. We know that the perimeter of a rectangle is given by the formula P = 2L + 2W. In this case, the perimeter is equal to 800 yards.
So we can write the equation as: 2L + 2W = 800.
Now, we need to express the area of the rectangle in terms of L and W. The formula for the area of a rectangle is A = L * W.
To find the maximum area, we need to find the values of L and W that satisfy the perimeter equation and maximize the area.
One way to solve this problem is to express one variable in terms of the other and substitute it into the area equation. Let's solve the perimeter equation for one of the variables (W) and substitute it into the area equation:
2L + 2W = 800 (divide both sides by 2)
L + W = 400 (subtract W from both sides)
L = 400 - W
Substitute L = 400 - W into the area equation:
A = L * W
A = (400 - W) * W
A = 400W - W^2
Now, we have the area equation A = 400W - W^2. To find the maximum area, we need to find the value of W that maximizes the equation.
One way to do this is by finding the vertex of the quadratic equation. The vertex can be found using the formula x = -b / (2a), in this case, W = -400 / (2*(-1)) = 200.
So, when W = 200, the area is maximized.
Now we can substitute the value of W into the expression we found for L:
L = 400 - W
L = 400 - 200
L = 200
Therefore, the dimensions of the rectangle that maximize the enclosed area are 200 yards for both length and width.
To find the maximum area, we can substitute the values of L and W into the area equation:
A = L * W
A = 200 * 200
A = 40,000 square yards.
Therefore, the maximum area is 40,000 square yards.
A square will provide the maximum area.
800 / 4 = _______ yards on each side