Maximum Area Question

[motlehh]

Hey..I'm new here. Need some help for a test tomorrow! Here's the problem:

If 570 yards of fence are used to enclose a rectangular plot of land using an existing wall as one side of the plot, write a function representing A(x). Find the dimensions of the rectangle with maximum enclosed area.

I already know the function would be f(x)=x(570-2x), but I don't understand how to find the max area.

It also asks a follow-up question of "what are all possible values?". I don't know how to do that either.

Any help would be appreciated. Thank you!

 

[tkhunny]

All Possible Values

x > 0 or it isn't rectangular or doesn't exist.

570 - 2x > 0
570 > 2x
x < 285 or is isn't rectangular or doesn't exist.

Didn't the problem statement say to use A(x)? Read carefully. That will help.

A(x) = x*(570-2x)

This is a simple quadratic function. Why do you not know how to find it's maximum value? You should not need calculus for that. Remember all that work with parabolas back in Algebra I, Algebra II, College Algebra, and Analytic Geometry?

With calculus, the derivative will help you. Find A'(x).

 

[TchrWill]

If 570 yards of fence are used to enclose a rectangular plot of land using an existing wall as one side of the plot, write a function representing A(x). Find the dimensions of the rectangle with maximum enclosed area.

I already know the function would be f(x)=x(570-2x), but I don't understand how to find the max area.

It also asks a follow-up question of "what are all possible values?". I don't know how to do that either.

The maximum enclosed area given a fixed perimeter is a square.

If one side is represented by a wall, fence, line, etc. the maximum enclosed area given a fixed perimeter is given by x by 2x where x + x + 2x = p, the given perimeter.

Considering all rectangles with the same perimeter, the square encloses the greatest area.
Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

Apply the same logic to your rectangle problem with one side already defined.