**Answer 1**

No calculus is really needed, but I'll give that solution after this one.

Letting the vertical sides be x and the horizontal ones for the small rectangles be y, we know that 3x+4y=200. Applying AM-GM to 3x and 4y, we get

(3x+4y)/2>=rt(12xy)

10000/12>xy

So the maximum area is 10000/12=2500/3, and this occurs when 3x=4y. Plugging this into 3x+4y=200, we get y=25 and x=100/3.

Now the calculus solution: from 3x+4y=200, we wish to maximize xy. Plugging in y=50-3x/4, we get

-3x^2/4+50x

this is maximized by taking the derivative and setting it equal to 0, so

-3x/2+50=0

x=100/3

Plugging this in, we find that y=25 as we did before.

**Answer 2**There are 3 "vertical" lines of length x and 4 "horizontal" lines of length y (2 above and 2 below)

So 3x+4y = 200 and the sum of areas is 2xy which is required to be maximum. Now eliminate y.

So f(x) = 2x(200 -3x)/4 is rquired to be max.

So f ' (x) = 100 - 3x =0 or x = 100/3, for stationary values of f(x)

and f"(x) = -3 <0 so f is max when x = 100/3. QED