Question

A rancher has 1200 feet of fencing to construct six​ corrals, as shown in the figure. Find the dimensions that maximize the enclosed area. What is the maximum​ area?

Answer 1

To maximize the enclosed area with 1200 feet of fencing, calculus optimization techniques should be employed. By modeling the corral as a rectangle and setting up equations for the perimeter and area, one can solve for the dimensions that maximize the area. Without a provided figure, exact dimensions cannot be given.

Step-by-step explanation:

To find the dimensions that maximize the enclosed area of the corrals with 1200 feet of fencing, we need to use optimization techniques that are generally covered in high school mathematics, specifically in calculus.

We can model the situation with a rectangle that is subdivided into six smaller rectangles (corrals) by five equally spaced divisions. Given the perimeter P of the outer rectangle (1200 feet in this case), and labeling the length of the rectangle as L and the width as W, the total perimeter of the fencing used would be given by P = 2L + 6W.

To maximize the area A, which is A = L * W, we set up the equation using the given perimeter and solve for one of the variables in terms of the other.

First, we express L in terms of W using the perimeter constraint: L = (P - 6W) / 2. The area equation becomes A = [(P - 6W)/2] * W. We then take the derivative of A with respect to W, set it equal to zero, and solve for W to find the width that maximizes the area.

Substituting this value back into the perimeter equation gives us the corresponding length L that maximizes the area.

However, without additional details and a diagram, we can't provide the exact numerical solution. Still, these steps outline the process for a maximization problem involving area and fencing. Without the ability to visualize the given figure, the answer won't be complete.

Answer 2

Step-by-step explanation:

To maximize the enclosed area, we need to divide the 2300 feet of fencing equally between the six corrals. Thus, each corral will have 2300/6 = 383.33 feet of fencing.

Let x be the length of each rectangular corral and y be the width. Then, we have the following equations:

2x + y = 383.33 (fencing equation)

A = xy (area equation)

We can solve the first equation for y:

y = 383.33 - 2x

Substituting this expression for y into the area equation, we get:

A = x(383.33 - 2x)

Expanding and simplifying, we have:

A = 383.33x - 2x^2

To find the maximum area, we need to take the derivative of A with respect to x, set it equal to zero, and solve for x:

dA/dx = 383.33 - 4x = 0

4x = 383.33

x = 95.83

Substituting x = 95.83 back into the fencing equation, we get:

2(95.83) + y = 383.33

y = 191.67

Therefore, the dimensions that maximize the enclosed area are 95.83 feet by 191.67 feet, and the maximum area is:

A = 95.83 * 191.67 = 18386.46 square feet.