Optimization: A rancher has 200 feet of fencing with which

[paulxzt]

A rancher has 200 feet of fencing with which to enclose in two adjacent rectangular corrals. What dimension should be used so that the enclosed area will be a maximum?

The figure is given as two adjacent rectangles side by side and the bottom is
<--x--> for the first rectangle and <---x---> for the second. The sides are just y.

Can someone show me the steps to getting the lengths? I get x = 25 but also get y = 50 but the answer is given as 25 and 33.33 somehow? Am i doing something ridiculously stupid?

 

[stapel]

How did you get your answers?

Please be specific. Thank you.

Eliz.

 

[paulxzt]

Primary:
P = 4x + 2y

Secondary:
A = 2xy

200 = 4x + 2y

100 - 2x = y

A = 2x (100 - 2x)

200x - 4x^2

A' = -8x + 200 = 0
x = 25.
y = 50

I probably made a stupid mistake. help is appreciated.

 

[stapel]

Primary:
P = 4x + 2y

How did you get four lengthwise pieces, when the area is divided into only two corrals (and thus there should be only three lengthwise pieces: two down the sides, and one down the middle).

I may be misunderstanding the exercise; you are, after all, looking at a picture that I can't see. Please clarify.

Thank you.

Eliz.

 

[paulxzt]

Primary:
P = 4x + 2y

How did you get four lengthwise pieces, when the area is divided into only two corrals (and thus there should be only three lengthwise pieces: two down the sides, and one down the middle).

I may be misunderstanding the exercise; you are, after all, looking at a picture that I can't see. Please clarify.

Thank you.

Eliz.

So would it be 2x + 3y? I mean, the figure has <--x--> below the first rectangle and <--x--> below the second so wouldn't i just call that 2x and 2x of the other side?

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<-----x----> <---x------->[/code]