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