# Thread: Optimization Problem

1. ## Optimization Problem

this is a rectangle area optimization problem:

A farmer wants to enclose a rentangular area with a fence. He would like the fence to be as inexpensive as possible. Assume the fencing material cost $1 per foot. a) suppose$40 is available for the project, what is the maximum area that can be enclosed.

b) suppose 100 square feet must be enclosed. what is the least possible cost?

2. The maximum area of a rectangle with fixed perimeter is a square.

Therefore, the maximum area of the enclosure with a 40 foot perimeter can be found by:

$\displaystyle \left(\frac {40}{4}\right)^2$

3. To find both of these with calculus:

We know that 40 feet of fencing is available. With this knowledge, find an equation for the area that is dependent on one side of the fence:

$\displaystyle A=lw$

$\displaystyle 2l=40-2w$

$\displaystyle l=20-w$

$\displaystyle A(w)=w(20-w)=20w-w^2$

Find the derivative:

$\displaystyle A'(w)=20-2w$

Find the critical points. These occur when the equation is either undefined or equal to 0. Since this equation is never undefined, find where it is equal to 0:

$\displaystyle 20-2w=0$

$\displaystyle 20=2w$

$\displaystyle w=10$

Using the formula $\displaystyle l=20-w$ that we found earlier, we can see that the maximum area is when the enclosure is 10 by 10.

Therefore, the maximum area is 100 square feet.

To find the least possible cost to make a 100 square foot enclosure, use the same process. We know that the cost is equal to the perimeter so we must find the smallest perimeter possible. First find an equation for the perimeter in terms of the width:

$\displaystyle P=2l+2w$

We know that the area is given by:

$\displaystyle A=lw$

$\displaystyle 100=lw$

$\displaystyle l=\frac {100}{w}$

Therefore:

$\displaystyle P=2(\frac {100}{w})+2w$

$\displaystyle P(w)=\frac {200}{w}+2w$

Now find the critical points. First we have to find the derivative:

$\displaystyle P'(w)=\frac {-200}{w^2}+2$

We can see that this equation is undefined when w is 0 but that is an extraneous solution in this scenario. So, when is this equal to 0?

$\displaystyle 0=\frac {-200}{w^2}+2$

$\displaystyle \frac {-200}{w^2}=-2$

$\displaystyle \frac {1}{100}=w^2$

$\displaystyle w=\pm \sqrt {\frac {1}{100}}$

We know that the width can not be negative, so our only answer is:

$\displaystyle w=\frac {1}{10}$

Can you solve it from here?