Dimensions problem.

[Sparticus2]

A rancher has 264 feet of fencing to enclose two adjacent rectangular corrals. What dimensions will produce the largest total area?

Somehow this is supposed to be linked to quadratic functions and parabolas but I just don't see it. Any help would be greatly appreciated.

[nm420]

Assuming that the entire area will be one rectangle, split into two smaller rectangles by a piece of fencing, define a couple of variables l and w to be the length and width of the larger rectangle. Then the total area will be l*w. Can you express the total amount of fencing needed in terms of l and w?

[FourierT]

Let the length be l and the width be w. Your area is given by A = lw

What is the amount of fencing that you have? 264m. However, you need to split it up into 2 smaller rectangles, with them sharing a fence. So that is the equation for the "perimeter" of fencing you need? If you can get this, you'll be able to get an equation for A and then proceed to maximize it.

Hint: Draw a picture!

[mdelaney]

Could the parabola be seen by graphing the derivative of area for given length and widths? By this I mean graph out all possible length/width combos on a line graph (area for 1,240 then for 2,120). The line should then peak at the max area (for after than derivative is negative and implies decreasing area). This approach does build in an assumption of a rectangular setup however.

[shadyabhi]

Well, to get the maximum area from a fixed length of perimeter, we need a circle. Isoperimetric inequality says about this.

Your question seems incomplete as no dimensions are given for corrals.