A rectangle is inscribed between two parabolas y=4x^2 and y=30-x^2. What is the area of the largest rectangle that could be inscribed?
Please Help! I have no clue.
- δοτζοLv 71 decade agoFavorite Answer
Find where they intersect to find the bounds on the width
4x² = 30 - x²
5x² = 30
x² = 6
x = ±√6
So the width must be between 0 and 2√6 (exclusive) and is given by the function 2x. The height of the rectangle is given by the difference of the functions, larger minus smaller.
h = 30 - x² - 4x² = 30 - 5x²
A = hw = (30-5x²)(2x) = 60x - 10x³
A' = 60 - 30x² = 0 ⇒ x = ±√2 ⇒ x = √2
w = 2√2
h = 20
A = 40√2 ≈ 56.569
The key to this whole thing is graphing the 2 equations and trying to visualize an inscribed rectangle.