Question 985275: The Equation of a Parabolic Curve is given by y = -x^2 + 12. What is the Maximum Area of a Rectangle inscribed in the Parabola?. I am asked to provide a Geometrical solution or Algebraic Solution. I am working for hours to find a Solution using coordinate geometry but would appreciate it if I can get help.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! I'm assuming this rectangle has one edge on the x axis. Let P be one point of the rectangle. Also, let P be on the positive x axis to the left of the root of the given parabola.
The distance from (0,0) to P is some unknown quantity x. Due to symmetry, the edge of the rectangle that rests on the x axis is going to be 2*x units long.
The height of this rectangle will be -x^2+12 because the upper corner point is on the parabola (draw a picture and this should be clear hopefully).
So we have length = 2x and width = -x^2 + 12
The area of this rectangle is...
A = L*W
A = (2x)*(-x^2 + 12)
A = -2x^3 + 24x
There are two ways to find the max area
1) Use a graphing calculator to find the peak max point on y = -2x^3 + 24x. This is like the vertex of a parabola, but not entirely. I like to think of it like it though. If you have a TI calculator, you can use the min/max feature to find the peak point.
2) Use calculus to find the max without using a graphing calculator. This is only if you're in a calculus class.
Using either method (I recommend method #1 since it sounds like you're in a high school algebra class), you should find that the max occurs at the point (2,32).
So if x = 2, the length is 2*x = 2*2 = 4 and the width is -x^2 + 12 = -(2)^2 + 12 = 8. Notice how L*W = 4*8 = 32
The max area possible is 32 square units (length = 4, width = 8)
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