Question

The land is shaped like a rectangle with sides of 20 meters and 30 meters. the owner intends to surround it with a fence and divide it into two parts with the same fence, one of which is in the shape of a square. determine the maximum area.

Answer 1

To find the maximum area of the rectangular land divided into two parts with a square fence, we can use an equation and calculus. By finding the value of x that maximizes the equation, we can determine the maximum area.

Step-by-step explanation:

To determine the maximum area of the rectangular land, we need to divide it into two parts with the same fence. One part is in the shape of a square. Let's call the length of the square side x. The remaining part of the rectangle will have dimensions of 20 meters by (30 - x) meters. The area of the square is x^2 and the area of the rectangle is 20(30 - x). Since the total area is the sum of the square and the rectangle, we can write the equation: x^2 + 20(30 - x) = maximum area. To find the maximum area, we can find the value of x that maximizes this equation by taking the derivative and setting it equal to zero. After solving for x, we can substitute the value back into the equation to find the maximum area.

Answer 2

The student's question involves applying principles of geometry and algebra to maximize the area that can be fenced, including a square. By setting up an equation based on the total available fencing and perimeter, the side length of the square can be found, and the maximum area can be calculated through differentiation.

Step-by-step explanation:

The student is asking how to determine the maximum area that can be enclosed by a fence on a plot of land with the shape of a rectangle, with one of the divided parts being a square. To find the maximum area that can be enclosed, we first need to recognize that by dividing the rectangle into a square and another rectangle, the sum of the perimeters of the two shapes should be equal to the total amount of fence available.

Let's assume the side of the square is x meters. The other rectangle would then have dimensions of 20 meters by (30 - x) meters. The total perimeter would be 2x + 2(20) + 2(30 - x) + x = 100, as we have 100 meters of fencing (2x for the sides of the square, 20 meters for one side of the rectangle, and 30 meters for the other side of the rectangle).

Solving this equation for x gives us the side length of the square, and thus we can calculate the maximum area for the square as x2. To find the maximum area, we should differentiate the area with respect to x and find the maximum value. The area for the other part of the land (the rectangle) is 20 * (30 - x), and the combined area should be maximized