Hi guys,
I am really stuck with this problem, maybe somebody could help me out.
A rancher has 2100 feet of fencing with which to construct adjacent, equally sized rectangular pens as shown in the figure above. What dimensions should these pens have to maximize the enclosed area?
This is a picture that belonged to it as well.
i(DOT)imgur(DOT)com/qh13W07.jpg
x=
y=
Maximum area=
Thanks for your help!
Frank
I could not get the image to come up, but you will find that the maximum area is achieved when it is equally split among the widths and lengths.
So, 1050 will be divided among the vertical pieces, and 1050 will be split into horizontal pieces.
Define the 2100 as the sum of *x + *y = 2100
and define x or y in terms of the other.
Then you can express the area as a function of just x or y.
It will be a parabola, and the maximum area is at the vertex.
My Progress so far:
P=4x+3y=2100
A=2xy
A=2x(2100-4x/3) This way I was able to calculate x= 525/2 or 262.5 But I'm having a struggle rewriting the formula used in terms of y. Can somebody help?
Hi Frank!
To find the dimensions that will maximize the enclosed area, we need to maximize the area of each pen.
Let's start by defining the dimensions of each pen. Since they are adjacent and equally sized rectangular pens, let's say the length of each pen is x feet.
Now, we need to determine the width of each pen. Since adjacent pens share a common fence, we need to subtract the length of one fence from the total perimeter. There are three sides that form the perimeter of each pen: the length (x), the width (y), and the common fence (which we'll call h).
The total perimeter equation is:
Perimeter = 2x + y + h = 2100 feet.
Now, we know that the common fence h is split evenly between the two pens, so h/2 is the width of each common fence that separates the pens. Also, since adjacent pens share a common fence, we know that the combined width of the two pens is y + h/2.
So, the width of each pen is y + h/2.
Now, let's substitute these values into the perimeter equation:
2x + y + h = 2100.
Since h/2 is the width of the common fence, we can rewrite the equation as:
2x + y + (2 * h/2) = 2100.
Simplifying this equation gives us:
2x + y + h = 2100.
Now, let's isolate y in terms of x and h:
y = 2100 - 2x - h.
To maximize the enclosed area, we need to find the maximum value for the area function. The formula for the area of a rectangle is:
Area = length * width.
For each pen, the length is x and the width is y + h/2. So, the area of each pen is:
Area = x * (y + h/2).
Now, let's substitute the equation for y into the area equation:
Area = x * ((2100 - 2x - h) + h/2).
Simplifying this equation gives us:
Area = x * (2100 - 2x + h/2).
Now, we have an equation for the area in terms of x and h.
To maximize the area, we need to find the values of x and h that satisfy the perimeter equation and give us the maximum value for the area equation.
Unfortunately, without specific values for x and h in the problem or the given picture, I am unable to provide you with the exact dimensions and maximum area.
However, you can solve the system of equations formed by the perimeter and area equations to find the values of x, y, and the maximum area. You can use algebraic methods such as substitution or elimination to solve the equations.
Once you have the dimensions of the pens in terms of x and y, you can find the maximum area by substituting those values into the area equation.