Maximizing Enclosed Area with 400ft Fencing

[hankjr]

Homework Statement

A person has 400 ft. of fencing to enclose two adjacent rectangular regions of the same size. What dimensions should each region be so that the enclosed area will be a maximum.

Homework Equations

The Attempt at a Solution

I have no Idea how to solve this problem, please please help me !

 

[berkeman]

Sounds like you need to build a fence with a 400 ft. perimeter, and maximize the enclosed area. Then just cut the internal area in half to see what the resulting rectangles are. Write the equation for the area of a 4-sided figure (rectangle/square) as a function of its perimeter. Then do you know how to maximize the area as a function of the side dimensions?

 

[hankjr]

I'm not sure how to write the area as a function of its perimeter?

 

[Kurdt]

The following thread covered a similar question and may be of use to you.

https://www.physicsforums.com/showthread.php?t=146884

 

[dontdisturbmycircles]

Are you supposed to find the max with a first-derivative test or by deduction through transformations? i.e "vertex form"

 

[dontdisturbmycircles]

I never heard of the h = -b/(2a) formula to find the axis of symmetry although of course it makes sense now that I think of it. That may come in handy later on. Heh

 

[z-component]

Since this is posted in the calculus section, I'm guessing the OP needs to use the first derivative test. Start by generating one equation for perimeter and one for area. Since you need to maximize area, you need to work with the area equation, so the perimeter needs to be combined into the area equation using substitution. Once you have the final area equation in terms of one variable, you may apply the first derivative test to find the critical values (when the derivative equals 0 or is undefined).