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Algebra
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A farmer has 1600 ft of fence and wishes to build a rectangular enclosure that has inner fences so that the area look likes a 2 by 3 table. As a result there will be 6 smaller enclosures within the larger rectangle. Find the dimensions of the largest rectangular enclosure that can be built with the amount of fence available. Also find the area.

Oct 9th, 2014

Fence Total = Outside Perimeter +inside perimeter

Fence Total= 2L + 2W + L + 3W 

Fence Total=3L + 5W

Since W gets factored 2 more times than L, to maximize area we will make sure the W is the smaller side of the rectangle while the L is the larger side. This is usually the case anyways when we name sides of a rectangle. 

Area= LW 

We want LW to be the biggest number such that 1600=3L + 5W

Solve for L and substitute into LW

(1600-5W)/3 =L

Area= (1600W-5W^2)/3

We want area to be maximized. We have a quadratic we can find the maxima of a quadratic using algebra by doing (c-b^2/4a)

This formula is explained here: http://www.wikihow.com/Find-the-Maximum-or-Minimum-Value-of-a-Quadratic-Function-Easily

Using algebra... 0- ((1600/3)^2)/(-20/3) =  42666.67

 or we can use calculus which is much more intuitive and doesn't require you to know arcane formulas!

Lets take the derivative...

d/dx Area = 1600/3- 10W/3

Area is a constant so...

0=1600/3 - 10W/3

1600/3 = 10W/3

W=160

If W is 160, then L=(1600-5(160))/3=800/3 = 166.67 

Therefore, max area= 160* 800/3 = 42666.67

Oct 9th, 2014

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