Apologies for my English, I'm a not a native speaker...
So I've got this homework, about finding the maximum area of a rectangle under a parabola. I'm using this as a reference to do my work: How to find the dimensions of a rectangle if its area is to be a maximum?
After searching for similar questions, the problems I've found are using parabolas symmetric with respect to the $Y$-axis, thus the rectangle in question can be "split" in half with the same width to the left and to the right of the $Y$-axis, and then we can take $(-x,0)$, $(x,0)$, $(x,y)$ and $(-x,y)$ as the rectangle's points with $(x,y)$ and $(-x,y)$ touching the parabola and $(-x,0)$ and $(x,0)$ touching the x-axis (like my reference above). Could I still do this if the parabola isn't symmetrical with respect to the $Y$-axis? Because then the rectangle wouldn't be "split in half" anymore...
I've tried to do that anyway and I got a good answer.
- The parabola's equation is $f(x)=-x^2+4x+3$, from this I determined the rough shape of the parabola (by determining the highest point of the graph which is $(2,7)$ and the y-intercept which is $y=3$), and placed the rough location of the rectangle inside the parabola
The graph: http://i.imgur.com/M0Hc4JU.png
- The width of the rectangle is $2x$, the height is $y$. So the area is $2xy$ (I got all these from my reference above, which uses a parabola symmetrical with respect to the $Y$-axis. I'm not sure if I can still use $2xy$ because my parabola isn't symmetrical with respect to the $Y$-axis, hence the question)
- Substituting $f(x)=y$ into $2xy$ I got $A(x)=-2x^3+8x^2+6x$
- $A'(x)=-6x^2+16x+6$ with its roots $3$ and $-(1/3)$
- Substituting $x=3$ into $A(x)$ I got $36$ as the maximum area of my rectangle
Thank you for the kind help