First you have to find their intersection
which is an easy equation to solve
4x^2 = 30-x^2
x = ± √6
y = 24
so area = integral (30 - 5x^2) from -√6 to √6
or
= integral 2(30-5x^2) from 0 to √6
= 60x - (5/3)x^3 | from 0 to √6
= 60√6 - (5/3)6√6 - 0
= 50√6
A rectangle is insribed between the parabolas y=4(x^2) and y=30-x^2. What is the maximum area of the rectangle?
Domain: [-3,3]
Range:[-2, 40]
THANK YOU!
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