optimization?

[tanja]

If you are given a downfacing parabola with its vertex on the y axis, such as y= 4-x^2, and also given a line such as y=x+2, what is the maximum area of a rectangle that can be formed between the 2 equations and the x axis?

[dessskris]

I don't really get your question. which area are you referring to? the x-axis is not lying on the area bounded by the two graphs

anyway, in this type of question, you're supposed to find the width and height of the rectangle in terms of x. for the height, look at the function of the graph above the rectangle. then, find the area i.e. width times height, and differentiate to find the maximum area.

[Keel]

The x-axis acts as another boundary.

But the above approach is correct. You must first figure out which function you wish to use to determine the height, its easier to use the linear one.

So you start of with a value, x₁, which is between -2 and 1 due to limits of height and the function.

The height at this given point, y, is given by the the chosen function, y=x+2. Thus y=x₁+2.

To find the width you will need to use the above to find x₂. It can be seen that x₁ = y-2 = -x₂²+2.

Now you can write everything in terms of x₁:

The width is given by x₂-x₁: sqrt(2-x₁) - x₁

The height is given by: x₁+2

Thus the area is the product of both. Proceed then to differentiate and find the value for x₁ which corresponds to the maximum area and plug that back into the Area equation.