Asked by Jim
A rectangle is inscribed between the parabolas y=4x^2 and y=30-x^2. what is the maximum area of such a rectangle?
a)20root2
b)40
c)30root2
d)50
e)40root2
a)20root2
b)40
c)30root2
d)50
e)40root2
Answers
Answered by
Reiny
one parabola opens up , the other down, so you have an enclosed region.
you will need their intersection
4x^2 = 30-x^2
5x^2 = 30
x = ± √6
height of the region = 30-x^2 - 4x^2 = 30-5x^2
because of symmetry we can take the area from 0 to √6 and double it
area = 2 integral [30-5x^2] from 0 to √6
can you take it from there?
let me know what you got.
you will need their intersection
4x^2 = 30-x^2
5x^2 = 30
x = ± √6
height of the region = 30-x^2 - 4x^2 = 30-5x^2
because of symmetry we can take the area from 0 to √6 and double it
area = 2 integral [30-5x^2] from 0 to √6
can you take it from there?
let me know what you got.
Answered by
Anonymous
Uhh still don’t get it
Answered by
Linlin
Using Maxima and Minima:
Given:
y = 4x² (upward)
y = 30 - x² (downward)
A = lw
A = (2x)(30 - x² - 4x²)
A = 2x(30 - 5x²)
A = 60x - 10x³
A' = 60 - 30x² = 0
30x² = 60
x² = 2
x = +√2
w = 2√2
l = 30 - 5x² = 30-10 = 20
A(max) = lw = (2√2)*20 = 40√2
Answer: e
Given:
y = 4x² (upward)
y = 30 - x² (downward)
A = lw
A = (2x)(30 - x² - 4x²)
A = 2x(30 - 5x²)
A = 60x - 10x³
A' = 60 - 30x² = 0
30x² = 60
x² = 2
x = +√2
w = 2√2
l = 30 - 5x² = 30-10 = 20
A(max) = lw = (2√2)*20 = 40√2
Answer: e
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