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April 18, 2015

April 18, 2015

Posted by **Gabriela** on Monday, December 13, 2010 at 4:31pm.

- calculus -
**Reiny**, Monday, December 13, 2010 at 5:21pmlet the point be P(x,y)

Perimeter = 2x + 2y

= 2x + 2(x^-2)

= 2x + 2/x^2

d(perimeter)/dx = 2 -4/x^3

= 0 for min of perimeter

2 - 4/x^3 = 0

2 = 4/x^3

x^3 = 2

x = 2^(1/3) , (the cube root of 2)

y = 2^(-2/3)

(I really expected the point to be (1,1) but the Calculus shows my intuition was wrong

the perimeter with the above answers is 3.78

had it been (1,1) the perimeter would have been 4)

- calculus -
**Rob**, Sunday, April 7, 2013 at 1:14pmTry this way:

Perimeter = 2x+2

therefore P = 2x + 2/x^2

d(perimeter)/dx = (2x^3-4)/x^3

solving for x = 2^(1/3)

therefor y was given as y = x^-2

plug x into y to solve for y.

y = 2^(-2/3) or y = (1/4)^(1/3)

check the answer it should be 3.78

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