I usually know how to do these types of problems, but the second variable just threw me off balance..
47. A rancher has 200 feet of fencing to enclose two adjacent rectangular corrals. What dimensions will produce a maximum enclosed area? (the diagram is of two adjacent corrals sharing a middle fence, y, and each individual corral has length x [so the two together would be 2x for the entire length])
I hope I've explained the diagram enough...any help is greatly appreciated!! :D
Fencing needed= 3Width + 4 length, correct? I don't have your diagram
Area= 2*W*L
so 200=3w+4L so l= (200-3w)/4 putting that into area..
area= 2w(200-4w)/4
Ohhhhhh i totally get it now! Thanks!! :D
No worries! I can help you with that. To find the dimensions that will produce a maximum enclosed area, we need to use optimization techniques.
Let's start by breaking down the problem and identifying the variables involved. We have two adjacent rectangular corrals, and each individual corral has a length of x (so the total length for both corrals together would be 2x). The only given information is that the rancher has 200 feet of fencing available.
Now, let's create an equation to represent the problem. We know that the total length of the fencing required is 200 feet, which consists of the lengths of the three sides of the corrals.
For the two individual rectangular corrals, we have the following sides:
- Length: x
- Width: y
So, the total length of the fence used for the two individual corrals is 2x (for the length) + 2y (for the widths), which should equal 200 feet.
2x + 2y = 200
Next, let's express the area of the two corrals in terms of our variables x and y. The area of each corral is given by length multiplied by width. Since we have two corrals, the total enclosed area would be the sum of the two individual areas.
Area of each corral: x * y
Total enclosed area: 2 * (x * y) = 2xy
Now, let's solve the system of equations to find the values of x and y that maximize the enclosed area.
We have two equations:
1. 2x + 2y = 200
2. Total enclosed area (A) = 2xy
To proceed, we need to isolate one variable from the first equation and substitute it into the second equation. Let's solve the first equation for x:
2x + 2y = 200
2x = 200 - 2y
x = 100 - y
Now, substitute x = 100 - y into the second equation to get the total enclosed area in terms of y:
A = 2xy
A = 2(100 - y)y
A = 200y - 2y^2
To find the maximum area, we can take the derivative of A with respect to y and set it equal to zero:
dA/dy = 200 - 4y = 0
Solving this equation will give us the value of y that maximizes the area.
200 - 4y = 0
4y = 200
y = 50
Now, substitute the value of y back into x = 100 - y to find the corresponding x-value:
x = 100 - y
x = 100 - 50
x = 50
So, the dimensions that will produce a maximum enclosed area are x = 50 feet and y = 50 feet.