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.
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.
We want LW to be the biggest number such that 1600=3L + 5W
Solve for L and substitute into LW
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
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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