optimize this

[bradp]

Find the maximum area of a rectangle bounded by [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula].

[collegebookworm]

We're searching in the range [unparseable or potentially dangerous latex formula], because those are the x-value endpoints of the closed area between [unparseable or potentially dangerous latex formula] and [unparseable or potentially dangerous latex formula]. I'll define a new function, [unparseable or potentially dangerous latex formula] to make this optimization easier.

The area of a rectangle enclosed in the region is [unparseable or potentially dangerous latex formula]. To optimize the problem, one should look at the area of rectangles when x and y are at their endpoints and at their critical points.

The critical points of function A are where the derivative equals zero, so we have: [unparseable or potentially dangerous latex formula]. When the derivative is zero, [unparseable or potentially dangerous latex formula].

The value of A at the the endpoints and critical points are:

[unparseable or potentially dangerous latex formula]

Notice how the area for the endpoints is zero, which we would expect since the y-values at those x-values are zero. Notice too how the area for the [unparseable or potentially dangerous latex formula] x-value is exactly the negative of the area for the [unparseable or potentially dangerous latex formula] x-value, because [unparseable or potentially dangerous latex formula] is even. Thus, we can calculate the area at all more points on the curve, but we can safely say that the largest area is about [unparseable or potentially dangerous latex formula]

[bradp]

13.01 is adding the y axis as another bound, so the max rectangle area without the y axis as a bound would have twice the x coordinate, making the area 26.02

i think?

thanks, just needed to check my take home test :D

[collegebookworm]

oops, I'll be more careful next time :)

[AuSmith]

You could have shown that the maximum area rectangle has to have horizontal and vertical sides... sure, it's mostly intuitive since [unparseable or potentially dangerous latex formula] is so steep. But hey, no one ever asked you to prove anything on a math test - don't be silly.

Admittedly, I frown a little on this problem because if [unparseable or potentially dangerous latex formula] weren't so steep, it wouldn't be obvious that the maximum area rectangle was upright.

[bradp]

any even function should have its max rectangle as upright

don't get creative!

[AuSmith]

[bradp]

I knew you would catch me.

[AuSmith]

Then, once [unparseable or potentially dangerous latex formula] increases past [unparseable or potentially dangerous latex formula], I think the rectangle has to tilt because if it keeps its diagonals perpendicular, it's a rhombus. A rectangular rhombus is a square. And, if it's an untilted square, it can't touch all four corners to the boundary. All maximal area rectangles should touch every corner to the boundary, else you might push some of the other corners off, then stretch it some more.