# What are the dimensions of such a rectangle with the greatest possible area?A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2–x^2. I found the height...

What are the dimensions of such a rectangle with the greatest possible area?

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2–x^2. I found the height (4/3), but I can't find the width.

txmedteach | High School Teacher | (Level 3) Associate Educator

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Let's start from the beginning. The upper corners are going to be points on the part of the following parabola above the x-axis:

` f(x) = 2 - x^2`

Suppose that the point in the first quadrant defining the upper-right point of the rectangle will be `(x,y)`. We can then say the lower-right point will be `(x,0)`, the lower-left point will be `(-x, 0)`, and the upper-right point will be `(-x, y)`. A graph of the rectangle will look roughly like the following plot (I chose random numbers just to illustrate what I'm saying visually):

Our area to optimize will, therefore, be found by the following equation:

`A(x) = 2x * y`

Keep in mind, we can subsitute the function `f(x)` for `y` in the equation, giving us the following result:

`A(x) = 2x(2-x^2)`

Simplifying the equation:

`A(x) = 4x - 2x^3`

Now, like any other optimization problem, we will take the derivative and set it to zero to find our max area:

`(dA)/(dx) = 4 - 6x^2`

Now, let's solve for the roots to find the `x` at which the area is optimized:

`0 = 4-6x^2`

We can factor using a difference of squares:

`0 = (2-sqrt6 x)(2 + sqrt6 x)`

Solving this for the factor on the left by first subtracting 4:

`-2 = -sqrt6 x`

Now, divide by `-sqrt6`:

`2/sqrt6 = x`

Noting that the second solution will be the negative value of the above solution, we get two possible values for x:

`x = +- 2/sqrt6`

Let's concentrate on the positive value because we said the point that we will base our solution off of will be in the first quadrant (meaning `x>0`). We can now find the optimized area through either using the formula we derived above, or we can find the y-value to find the height of the rectangle. The first method just involves us substituting `2/sqrt6` into our area equation:

`A(2/sqrt6) = 4*2/sqrt6 - 2(2/sqrt6)^3 = 2.177`

So our maximum area should be 2.177 square units.

The second method simply involves us finding the `y`-value at `x = 2/sqrt6` using the parabola formula given in the problem:

`y = 2 - (2/sqrt6)^2 = 2 - 4/6 = 4/3`

There is the height you found! Now, let's incorporate it into finding the area by using `2x` for the width:

`A = 2x*y`

`A = 2*2/sqrt6*4/3 = 16/(3sqrt6) = 2.177`

Thankfully both ways yield the same result, further validating our answer!

I hope this helps!

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