Differentiation: Maximum Area (fencing corrals)

[2000WS6]

A rancher has 200 ft. of fencing with which to enclose two adjacent rectangular corrals (imageine two fecned rectangluar areas sharing a commmon boundary). What dimentions should be used so that the enclosed area will be a maximum?

So far I have:

Primary Equation: \(\displaystyle {A = xy}\) ---> \(\displaystyle {y =}\frac{200}{x}\)

*this is where I'm lost.... this is my best guess
Secondary Equation: \(\displaystyle {f(x) = 4x * 3}(\frac{200}{x})\)

Am I close? Please help. Thank you!

 

[mathnerd]

Re: Differentiation: Maximum Area

hi,

Perimeter = 2l + 3w for rectangle

so, solve for w.

A = lw

Plug in the w from perimeter into this area formula.

Then differentiate the area. Then set A` = 0. And solve for L. You should be able to get the dimensions then.

I hope this helped.

 

[stapel]

Primary Equation: \(\displaystyle {A = xy}\) ---> \(\displaystyle {y =}\frac{200}{x}\)

The fencing is the perimeter available, not the area available. The fencing encloses the area.

A rancher has 200 ft. of fencing with which to enclose two adjacent rectangular corrals (imageine two fecned rectangluar areas sharing a commmon boundary). What dimentions should be used so that the enclosed area will be a maximum?

Draw the picture:

+---------+---------+ ^
|         |         | |
|         |         | w
|         |         | |
+---------+---------+ v
<------L------------>

The perimeter is then 3w + 2L = 200. Solve this for one of the variables in terms of the other.

The area is Lw. Substitute from the step above to obtain an expression for the area in terms only of one variable. Then differentiate, etc. :wink:

Eliz.