Asked by Anonymous
find the the area of the largest rectangle that can be under the parabola y=9-x^2 above the x axis?
Answers
Answered by
oobleck
Let the base of the rectangle have length 2x.
Then the area of the rectangle is
a = 2xy = 2x(9-x^2)
da/dx = 6(3-x^2)
so da/dx = 0 when x = ±√3
we want x>0, so a has a max of 12√3
Then the area of the rectangle is
a = 2xy = 2x(9-x^2)
da/dx = 6(3-x^2)
so da/dx = 0 when x = ±√3
we want x>0, so a has a max of 12√3
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.