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.
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?
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
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
can I ask why did you subtract?
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