Question

Suppose a rectangular pasture is to be constructed using 1 2 linear mile of fencing. The pasture will have one divider parallel to one pair of sides and two dividers parallel to the other pair of sides, so that there are six congruent enclosures. What is the maximum total area of such a pasture, in square feet?

Answer 1

Answer:

$\displaystyle A=\frac{1}{192}$

Step-by-step explanation:

Maximization With Derivatives

Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.

We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.

The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.

The amount of fence needed to enclose the external and the internal divisions is

$P=4x+3y$

We know the total fencing is 1/2 miles long, thus

$\displaystyle 4x+3y=\frac{1}{2}$

Solving for x

$\displaystyle x=\frac{\frac{1}{2}-3y}{4}$

The total area of the pasture is

$A=x.y$

Substituting x

$\displaystyle A=\frac{\frac{1}{2}-3y}{4}.y$

$\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}$

Differentiating with respect to y

$\displaystyle A'=\frac{\frac{1}{2}-6y}{4}$

Equate to 0

$\displaystyle \frac{\frac{1}{2}-6y}{4}=0$

Solving for y

$\displaystyle y=\frac{1}{12}$

And also

$\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}$

Compute the second derivative

$\displaystyle A''=-\frac{3}{2}.$

Since it's always negative, the point is a maximum

Thus, the maximum area is

$\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}$