Area of Rectangle inside graph

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So I got the answer to this question but it seems a little iffy. If someone could just double check my work, that would be wonderful.

I am not sure if this matters, but the level of math in this question is AP Calc-AB.

Find the area of the largest rectangle (with sides parallel to the coordinate axis) that cna be inscribed in the region enclosed by the graphs $f(x)=18-x^2$ and $g(x)=2x^2-9$ .

there is a picture, but it isn't that helpful.

Attachment 9443

so this is what i did. since we are looking for the largest area, i started with the area formula. $A=bh$ . I labeled each the distance from the y axis x. and since the rectangle is touching each graph, these are the x coordinates. $x, 2x^2-9$ for the bottom right hand corner and $x, 18-x^2$ for the top right hand corner. So I know $b=2x$ and $h$ =the distance between $x, 2x^2-9$ and $x, 18-x^2$ . So i used the distance formula; $d=\sqrt{(x-x)^2+(18-x^2-2x^2+9)^2}$
$d^2=(18-x^2-2x^2+9)^2$
$d^2=(27-3x^2)^2$
$d=27-3x^2$

so then i did $A=(2x)(27-3x^2)$ i got the derivative and $A'=54-18x^2$ i found the zeros and one was negative so the only possibility was x=1.732.

so x=1.732 will optimize the rectangle, correct?

correct decimal value, but you're making this way too difficult.

base of rectangle = $2x$

height of rectangle = $f(x) - g(x) = 27 - 3x^2$

$A = 2x(27 - 3x^2) = 54x - 6x^3$

$\frac{dA}{dx} = 54 - 18x^2$

$54 - 18x^2 = 0$

$3 - x^2 = 0$

$x = \sqrt{3}$

now sub in $\sqrt{3}$ for x in the area equation and ATQ.

o wow, thanks you so much, that seems much easier.