SOLUTION: A rectangle is placed under the parabolic arch by f(x)=27-3x^2 by using a point (x,y) on the parabola,as shown on the figure.Write a formula for the function A(x) that gives the ar

Algebra ->  Rectangles -> SOLUTION: A rectangle is placed under the parabolic arch by f(x)=27-3x^2 by using a point (x,y) on the parabola,as shown on the figure.Write a formula for the function A(x) that gives the ar      Log On


   

Question 73176: A rectangle is placed under the parabolic arch by f(x)=27-3x^2 by using a point (x,y) on the parabola,as shown on the figure.Write a formula for the function A(x) that gives the area of the rectangle as a function of the x-coordinate of the point chosen.
answers are:
a.f(x)=6(27-3x^2)
b.f(x)=27x-3x^3
c.f(x)=54x-6x^3
d.f(x)=162x-6x^3
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle is placed under the parabolic arch by f(x)=27-3x^2 by using a point (x,y) on the parabola,as shown on the figure.Write a formula for the function A(x) that gives the area of the rectangle as a function of the x-coordinate of the point chosen.
answers are:
a.f(x)=6(27-3x^2)
b.f(x)=27x-3x^3
c.f(x)=54x-6x^3
d.f(x)=162x-6x^3
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I don't have the figure but you do.
The base of the rectangle is 2x because it goes x to the right of the y axis
and x to the left of the y axis.
The height of the rectangle is y which equals 27-3x^2
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The Area is base*height
A(x) = (2x)(27-3x^2)
A(x) = 54x -6x^3 which is answer "c".
cheers,
Stan H.