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

Math 
Reiny,
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(9x^2)
= 18x  2x^2
d(area)/dx = 18  4x
so 184x=0
x = ....
take it from here, let me know what you got 
oops Math 
Reiny,
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 
Math 
Damon,
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 = 9x^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 
Math 
brandon,
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?