Find the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing. (Assume that the length is greater than or equal to the width.)

shiuldn't it be p=4w+3L

And, note that this maximum area occurs when the fencing is equally split among lengths and widths.

To find the dimensions of the rectangular corral that produce the greatest possible enclosed area, we can use the concept of optimization.

Let's assume that the length of each pen is x feet. Since there are two pens of the same size, the total length of fencing required would be 2x.

Given that 2x = 300 feet (the total length of the fencing), we can solve for x by dividing both sides of the equation by 2:
x = 300 feet / 2 = 150 feet.

Now, we know that the length of each pen is 150 feet. Let's also assume that the width of each pen is y feet.

Since there are two pens, the total width of the corral would be 2y. However, since it is a rectangular corral split into two pens, the total length of each pen would be the same as the total width of the corral.

Therefore, the total enclosed area of the corral would be:
Area = Length * Width
Area = (2x) * (y)
Area = 150 feet * y

To find the maximum area, we need to differentiate the area equation and equate it to zero:
d(Area)/dy = 150 - 0
150 = 0

This means that the y-value which satisfies this equation would give us the maximum area.

Solving for y:
150 = 0
y = 0

Since y = 0 does not make sense in this context, it means that there is no maximum area. However, we can see that the length of each pen, which is 150 feet, will result in the maximum possible area.

Therefore, the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing are:
Length: 150 feet
Width: Variable/Not defined

Perimeter=3W+3L (makes two pens)

Area=WL

300=3W+2L
L=150-1.5W

Area=W(150-1.5W)
= 150W-1.5W^2

this is a parabola, in standard form
w^2-100w+Area/1.5=0
a=1 b=-100 c=area/1.5

maximum=-b/2a=100/2=50

http://www.wikihow.com/Find-the-Maximum-or-Minimum-Value-of-a-Quadratic-Function-Easily