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Find the dimensions of the rectangle with the smallest area that can be made with the base of the rectangle on the x-axis and the top two vertices on the parabola y = 8 - x^2, where y > 0. The variable x can only be in the range [1,2].

If the base is x - -x = 2x and height is 8 - x², the area is then the product a(x) = 2x(8-x²).

That is, the area is a(x) = 16x - 2x³.

The derivative of this is a'(x) = 16 - 6x². Setting this to 0 gives 16 = 6x², so x² = 8/3. This means that x = sqrt(8/3) (since x = -sqrt(8/3) is outside the domain). That can be multiplied by root(3/3), giving x = sqrt(24)/3. Since the second derivative is a"(x) = -12x, which is negative at x = sqrt(8/3), this value gives a maximum value.

We are looking for a minimum, however. Checking the endpoints gives a(1) = 14 and a(2) = 16.

Having done this, it can be seen that the smallest area is at x=1.

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