Calculus/Clac

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Expert: - 12/1/2009


Question
I have many questions like this one and was wondering if you knew how to go about it? Thanks a lot.

Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola. (Round your answers to the nearest hundredth.)
y = 6 - x2

Answer
That must be the parabola y = 6 - x².
So if the base is at x and one side, it is at -x on the other, so the base width is 2x.
The height is y, and that is given as 6 - x².
The area of the rectangle is base*height, or 2x(6-x²) = 12x - 2x³.

If we take A(x) = 12x - 2x³, then A'(x) = 12 - 4x².
Setting this to 0 gives gives 12 = 4x², or 3 = x², or √3 = x.

The area is found by putting that back into A and getting A(√3) = 12√3 - 2√3³.
Now the √3²=3, so we have 12√3 - 6√3 = 6√3.

Now √3 = 1.732050808, so 6√3 = 10.39230485.
Rounded to the nearest 100th, that is 10.39.  

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