Arectangle is constructed with its base on theâ x-axis and two of its vertices on the parabola yequals=2525minusâxsquared2. what are the dimensions of the rectangle with the maximumâ area? what is theâ area?

You should have drawn1 - x-axis and y-axis in light pencil.2 - graphed a down-facing parabola with the top of the frown on the y-axis at y = 2.  It should be crossing the x-axis at ±√2.  This should be in dark pencil or another color.3 - In dark pencil or a completely new color, draw a rectangle with one of the horizontal sides sitting on top of the x-axis and the other horizontal side touching the parabola at each of the top corners of the rectangle. The rectangle will have half of its base in the positive x-axis and the other half on the negative x-axis.  It should be split right down the middle by the y-axis.  So each half of the base we will say is "x" units long.  So the whole base is 2x units long (the x units to the right of the y-axis, and the x units to the left of the y-axis)  I so wish I could draw you this picture...   In the vertical direction, both vertical edges are the same length and we will call that y.   The area that we want to maximize has a width 2x long, and a height of y tall. So A = 2xy     This is the equation we want to maximize (take derivative and set it = 0), we call it the "primary equation", but we need it in one variable. This is where the "secondary equation" comes in.  We need to find a way to change the area formula to all x's or all y's. Since it is constrained to having its height limited by the parabola, we could use the fact that y=2 - x2 to make the area formula in only x's.   Substitute in place of the "y", "2 - x2" into the area formula. A = 2xy = 2x(2 - x2)   then simplify A = 4x - 2x3     NOW you are ready to take the deriv and set it = 0 dA/dx = 4 - 6x2       0 = 4 - 6x2   6x2 = 4    x2 = 4/6 or 2/3 So x = ±√(2/3) Width remember was 2x.   So the width is 2[√(2/3)]Height is y which is 2 - x2 = 2 - 2/3  =4/3

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