Asked by Gabriela
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.
Answers
Answered by
Reiny
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)
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)
Answered by
Rob
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.