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.

My work:

  • 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


The axis of your parabola is at $x=-b/2a=2.$ If you re-write your equation substituting $x=x'+2$ then the new parabola is symmetric with respect to the $Y$ axis. Since the transformation is a horizontal translation, it does not affect considerations about the area of rectangles under the curve.

  • $\begingroup$ Sorry, I don't understand, where is $x=x'+2$ coming from? And does that mean that the parabola has to be symmetrical with respect to the $Y$ axis for the equation $A(x)=2xy$ to work? $\endgroup$ – data4pass Dec 14 '15 at 13:22
  • $\begingroup$ The substitution $x=x'+2$ is a horizontal shift in the $XY$-plane chosen such that the $Y$ axis becomes the axis of the parabola. And yes, the equation $A(x)=2xy$ supposes symmetry, otherwise it is $A(x)=2(x-2)y$ or more generally $A(x)=2(x+b/2a)y.$ $\endgroup$ – Justpassingby Dec 14 '15 at 13:26
  • $\begingroup$ Hmm, I think I'm more interested on using $A(x)=2(x−2)y$. Is the equation $A(x)=2(x+b/2a)y$ universal i.e. can be used for similar problems with different parabolas? Would you please explain how do you get $A(x)=2(x+b/2a)y$? $\endgroup$ – data4pass Dec 14 '15 at 13:34
  • $\begingroup$ I am shifting $x$ over a distance equal to the $x$-coordinate of the axis of the parabola, which is $-b/2a.$ That holds universally for vertical parabolas. $\endgroup$ – Justpassingby Dec 14 '15 at 13:47
  • $\begingroup$ Addendum: of course, only those parabolas that hold any rectangles above the $X$-axis at all, i.e., $a<0$ and $b^2-4ac>0$. $\endgroup$ – Justpassingby Dec 14 '15 at 14:02

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