# Area of Rectangle inside graph

• Jan 1st 2009, 02:38 PM
OnMyWayToBeAMathProffesor
Area of Rectangle inside graph
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 $\displaystyle f(x)=18-x^2$ and $\displaystyle g(x)=2x^2-9$.

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

so then i did $\displaystyle A=(2x)(27-3x^2)$ i got the derivative and $\displaystyle A'=54-18x^2$ 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?
• Jan 1st 2009, 02:47 PM
skeeter
correct decimal value, but you're making this way too difficult.

base of rectangle = $\displaystyle 2x$

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

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

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

$\displaystyle 54 - 18x^2 = 0$

$\displaystyle 3 - x^2 = 0$

$\displaystyle x = \sqrt{3}$

now sub in $\displaystyle \sqrt{3}$ for x in the area equation and ATQ.
• Jan 1st 2009, 02:56 PM
OnMyWayToBeAMathProffesor
o wow, thanks you so much, that seems much easier.