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.
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!