Asked by J
A rectangle is inscribed in the interior section cut from the parabola x = 1/4 y2 by the line x=3. One side of the rectangle lies on the line. Fine the maximum area of such a rectangle.
Answers
Answered by
Reiny
Make a sketch.
Let the points on the parabola be P(x,y) and Q(x,-y)
Area = 2y(3-x)
= 2y(3-(1/4)y^2)
= 6y - (1/2)y^2
d(Area)/dy = 6 - y = 0 for a max of Area
y = 6
the max area is 6(6) - (1/2)(36)
= 18 square units
check my arithmetic
Let the points on the parabola be P(x,y) and Q(x,-y)
Area = 2y(3-x)
= 2y(3-(1/4)y^2)
= 6y - (1/2)y^2
d(Area)/dy = 6 - y = 0 for a max of Area
y = 6
the max area is 6(6) - (1/2)(36)
= 18 square units
check my arithmetic
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.