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# Calculus/Calculus: Optimization

Question
My question is: Among the rectangles with the area 400 meters squared, what are the dimensions of the one with the largest perimeter. (Show your work.)

I've gotten this far:
A=xy
P=2x+2y
A=400

y=400/x
P=2x+2(400/x)
P=2x+(800/x)

P'=(2x)'+(x'800 - 800'x)
P'=2+800
P'=802

I know that the largest perimeter is 802, so I know that the x and y values should be 1 and 400, but I can't get to those numbers using calculus, which I need to do in order to get full marks on my exam.

If you are talking about the largest perimeter, there is no such thing in this context because there is no upper bound. A little about the calculus bit first:

Differentiating P=2x+(800/x)= 2x + 800* (x^-1) wrt x on both sides,

you get dP/dx = 2 -800* (x^-2)
= 2 -800/ x^2

and then setting dP/dx =0 gives 2 -800/ x^2 =0 ,  ie  x=20, y=20

But the second order derivative of P which is equals to 1600/x^3 is definitely positive (since x must be >0), which means the combination of x=20, y=20 is actually a minimum.

You cited x and y values to be 1 and 400 respectively, have you considered 0.5 and 800, which gives a much larger perimeter of 1601? What about 0.25 and 1600, which will yield an even larger value of 3200.5? Hopefully you can recognize that there is no so called "maximum" value which can be spoken of.

Why so? Try graphing out P = 2x +(800/x) for x>0; you should see a minimum point curve bordered by the vertical axis (asymptote) and the P=2x oblique asymptote.

While there is a clear lower bound (hence the existence of a minimum point), the curve grows on both sides of that minimum point without restriction. Because of this, you will never be able to discover a maximum value of P.

Hope this detailed discussion clarifies matters. Peace.

Calculus

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