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I am using these as references:

  1. How to find the dimensions of a rectangle if its area is to be a maximum?

  2. Does the symmetry of a parabola in finding the maximum area of a rectangle under said parabola matter? (related)


Problem: Find the maximum area of a rectangle inside a parabola whose equation is $f(x)=-x^2+4x+3$

My Work:

  1. Determining 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 placing the rough location of the rectangle inside the parabola.

graph 1

  1. Shifting the parabola to the left so that the $Y$ axis becomes the axis of the parabola (i.e. $f(x)=f(x+(−b/2a))$). The shift results with a new equation of the parabola, which is $f(x)=-x^2+7$. Graphing it.

graph 2

  1. From reference 2, the area's equation is $A(x)=2(x-(-b/2a))(-x^2+7)$ (I suspect this is where my mistake lies) where $-b/2a = 2$, the equation turns out to be a cubic equation $A(x)=-2x^3+4x^2+14x-28$
  2. $A'(x)=0$ (to obtain the maximum $A(x)$), with roots $x=-2/3$ and $x=2$
  3. Substitute $x=2$ to $A(x)$, $A(2)=-2(8)+4(4)+28-28$ which turns out to be $0$

Thank you for the help...

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    $\begingroup$ Yep, in step 3 you needed to use the original $f(x)$. $\endgroup$ – Justpassingby Dec 14 '15 at 17:19
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    $\begingroup$ You should not mix the two parabolas, the original and the shifted one. Either use the original $f(x)$, or alternatively (as user298667 suggests) use the shifted $f(x)$ without subtracting any more axis shift (because the parabola is already symmetric). $\endgroup$ – Justpassingby Dec 14 '15 at 17:30
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    $\begingroup$ What results did you get? In the first case the maximum is reached at $2+\sqrt\frac73,$ in the second case at $x=\sqrt\frac73.$ $\endgroup$ – Justpassingby Dec 14 '15 at 18:10
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    $\begingroup$ It's $-2+\sqrt\frac73,$ sorry about the sign error. And that is equal to your second solution if you take the plus sign in front of the root. $\endgroup$ – Justpassingby Dec 14 '15 at 18:24
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    $\begingroup$ Sorry I take back what I said about the sign error, it is +2 but it is easy to make the mistake because the coefficient $a$ is negative. $\endgroup$ – Justpassingby Dec 14 '15 at 18:56
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After you shift the parabola, you should use new parameters in $f(x)=-x^2+7$. That means $-b/2a=0$.

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My mistake indeed lies on Step 3, where I either have to:

  1. Use the original equation of the parabola (which is $f(x)=-x^2+4x+3$), $A(x)=2(x−(−b/2a))(−x^2+4x+3)$, or
  2. Use the new equation of the parabola i.e. the equation you got after transforming/shifting the parabola, $A(x)=2x(−x^2+7)$

Both equations derives from the original equation $A(x)=2xy$ (from my first reference).

Instead of doing either steps, I used the new parabola's equation ($−x^2+7$) while still subtracting the axis shift ($(x−(−b/2a))$) despite the fact with my new equation the parabola is already symmetric.

Thanks to Justpassingby and ZQD for the help!

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