Optimization Area- Calc? A rancher has 200 feet of fencing with which to enclose two adjacent recatngular corrals. Use calculus to analytically determine the demensions that should be used so that the enclosed area would be a maximum. The picture given looks like this: _ _ | | | - - Jstu imagine two fully closed rectangulars connected at the middle. The length with the two dashes are labeled x and x and the shorter width is labeled y. I have no idea what my secondary equation would be. I'm thinking the primary equation would be A=L*W

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.

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