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?
width=?
height=?
could someone help me out in how to do this problem?
2006-12-01 13:43:03 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Given an x, you can get the height or the rectangle from the parabola eqn, and the width is just 2x. Thus area is a function of only x. Note that |x|<3 because that's where the parabola crosses the x axis.
Derive the area(x) and examine that function.
Area(x) = 2x*(9-x^2) , right?
multiply out, get d/dx of Area(x), set to zero, solve for x.
2006-12-01 13:49:38 · answer #1 · answered by modulo_function 7 · 0⤊ 0⤋
Alright, as the parabola is symmetric across the y-axis, we try to find the point on the parabola which will give the greatest possible area of a rectangle when it is set to the square's top-right corner (Other sides bounded by x and y axis)
The area of the rectangle = base*height
base=x
height=f(x)
Area(rect1)= xf(x)= 9x-x^3
We then take the derivative of this function, getting:
f'(x)=9-3x^2
We set this equal to zero to get the critical points of the function.
9-3x^2=0
9=3x^2
3=x^2
+-sqrt(3)=x
As we are only looking for first quadrant answers, we throw out the -sqrt(3) as a possible answer. As f(x) is a second degree polynomial with no first degree component, we know that the critical point will be the location at which the area is greatest. Therefore, the vertexes that give the greatest area will be [sqrt(3), f(sqrt(3))] and [-sqrt(3),f(-sqrt(3))].
So, we have the width as equal to 2sqrt(3) and the height as equal to f(sqrt(3)) or 6.
Solution:
Width = 2sqrt(3)
Height = 6
2006-12-01 13:58:27 · answer #2 · answered by Jason 2 · 0⤊ 0⤋
OK, first plot a simple picture of the graph. Then figure out an equation for the area. The simplest way is to work only in the first quadrant, knowing that you'll eventually multiply the x value by 2. So set the x value = x and the y value = 9-x^2. Set up and area equation to get A=x(9-x^2). Then simplify, take the derivative of it and find the relative maximum. That will give you the x value(which has to be multiplied by 2) and then you can get the y value by putting the x into the equation.
2006-12-01 13:51:08 · answer #3 · answered by Cornabled42 1 · 1⤊ 0⤋
Width of the rectangle = 2x
Height of the rectangle = 9-x^2
Area of rectangle = 2x (9-x^2)
A = 18x - 2x^3
A' = 18 - 6x^2
Set to zero and solve for x
x=sqrt(3)
Width = 2 sqrt(3)
Height = 6
2006-12-01 13:49:56 · answer #4 · answered by z_o_r_r_o 6 · 1⤊ 0⤋