Question

4. Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed in the region enclosed by the graphs of f(x) = 18 – x^2 and g(x) = 2x^2 – 9.

Answer 1

find the derivatives, then plug it in to the orginal questioin.then use the length to find the area as it is an optimization question

Answer 2

not the anti-derivative?

Answer 3

let the vertex in quadrant I be (x,y)
so the base of the rectangle is 2x
its height is (18-x^2) - (2x^2 - 9)
= 27 - 3x^2

Area = 2x(27-3x^2)
= 54x - 6x^3
d(Area)/dx = 54 - 18x^2
= 0 for a max of area

18x^2 = 54
x^2 = 3
x = ±√3
using x = √3
area = 54√3 - 18(3)
= 54(√3-1) or approx 39.53

Answer 4

can I ask why did you subtract?

Answer 5

suppose we take x=1 (within the rectangle)
What is the height at x=1 ?
f(1) = 18-1 = 17
g(1) = 2-9 = -7
so the actual height of the rectangle when x=1 is 17 - (-7) = 24

Did I not do a subtraction of f(x) - g(x) ?
notice by subbing in x=1 into the simplified subtraction answer of 27-3x^2
I get 27 - 3 = 24

Answer 6

In response to Reiny, at the end, wouldn't it be "area = (54(sqrt3) -6(sqrt3)^3)? And then you would get an answer of app. 62.354?

Answer 7

Why is the base of the rectangle 2x?