A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=9-x^2. What are the dimensions of such a rectangle with the greatest possible area?

Answer 1

let the point of contact in the first quadrant be (x,y)
then the base of the rectangle is 2x and its height is y
Area = 2xy
= 2x(9-x^2)
= 18x - 2x^2
d(area)/dx = 18 - 4x
so 18-4x=0
x = ....

take it from here, let me know what you got

Answer 2

Well, this is an upside down parabola (sheds water) with x intercepts at x = -3 and x = +3 and vertex at (0,9)
so lets do it just for positive x and double the base at the end.
Area = x y
where y = 9-x^2
A = 9x - x^3
dA/dx = 9 - 3 x^2
that is zero when x^2 = 3
or x = +/- sqrt 3 use + sqrt 3
so y = 9 - 3 = 6
Now we double the base because we only did the right half
base = 2 sqrt 3
height = 6

Answer 3

change
<< = 18x - 2x^2 >> to

= 18x - 2x^3

then d(area)/dx = 18 - 6x^2
= 0 for a max/min of area
6x^2 = 18
x^2 = 3
x = ±√3
and y = 6

same result as Damon

Answer 4

how do I write a written equation? A jacket cots 28.00 more than twice the cost of a pair of slacks. If the jacket costs 152.00, how much do th slacks cost?