optimization problem

[cher_m412]

Question:

Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed in the region bounded by the graphs of:

f(x) = 8 - 2x[sup:6t14yf1z]2[/sup:6t14yf1z] and g(x) = x[sup:6t14yf1z]2[/sup:6t14yf1z] - 4

I had those two equations as the secondary equations but i wasnt sure whether you had to use both or how to do that.
Also I had the primary equation as xy=max (area of rec.)

When I took the derivative I got a minimum instead, please help if you can!!
thanks

 

[galactus]

The area of the bounded region is \(\displaystyle 8-2x^{2}-(x^{2}-4)=-3x^{2}+12\)

The area of the rectangle is given by A=2xy

So, you have \(\displaystyle A=2x(-3x^{2}+12)\)

Differentiate, set to 0 and solve for x.

Take notice, that the difference in the two quadratics creates another quadratic in which you can inscribe the rectangle. In your two originals, the rectangle will drop below the x-axis, but the dimensions will be the same.
If you do them separately, you get the same result by adding them up.

From the diagrams, the areas of the rectangles are the same.