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

Find the point P in the first quadrant on the curve y=x^-2 such that a rectangle with sides on the coordinate axes and a vertex at P has the smallest possible perimeter.

- calculus -
**Reiny**, Monday, December 13, 2010 at 5:21pm
let 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 1:14pm
Try 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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