How to find the width and height of a constrained rectangle? (Precalculus)

Question

Of all rectangles with base on the $x$-axis and upper corners on the parabola $y=5−x^2$, what are the dimensions (width and height) of the rectangle that has the greatest possible area?

I have no idea how to solve this. I tried graphing $y=0$ and the parabola, but then I don't know where to put the vertical lines. How do I approach this question?

Answer 1

Actually, there is a solution for this problem that does not use calculus, but instead uses the AM-GM inequality.

Let the bottom right corner of the rectangle be the point $(x, 0)$, then the area of the rectangle is

$ A(x) = 2 x (5 - x^2) $

This can be factored as

$ A(x) = 2 x (\sqrt{5} - x) ( \sqrt{5} + x ) $

So, we want to maximize $ x (\sqrt{5} - x) ( \sqrt{5} + x ) $

Let $F = a x , G = b (\sqrt{5} - x) , H = \sqrt{5} + x $

Assuming $a,b$ are positive, we want to maximize the product FGH.

Now we know from $AM-GM$ inequality that

$ \left( \dfrac{F + G + H}{3} \right) \ge \sqrt[3]{ FGH } $

Equality occurs when $F= G = H$ and this is the optimum point. Now, the idea is to make $F+G+H$ a constant, and to impose on the parameters $a,b$ the condition that $x$ obtained from $F=G$ is equal to $x$ obtained from $F = H$.

This means that the coefficient of $x$ in $F+G+H$ is zero, i.e.

$ a - b + 1 = 0 $

And, from $F = G$ , we obtain,

$ x = \dfrac{b \sqrt{5}}{a + b} $

And, from $F = H$, we obtain,

$ x = \dfrac{\sqrt{5}}{ a - 1} $

So the second equation relating $a$ and $b$ is

$ \dfrac{b \sqrt{5}}{a + b} = \dfrac{\sqrt{5}}{a-1} $

which simplifies to

$ b (a - 1) = a + b $

We need to solve for $a$ and $b$ from these two equations. From the first equation, we have

$ b = a + 1 $

Substituting into the second equation,

$ a^2 - 1 = 2 a + 1 $

so that

$ a^2 - 2 a - 2 = 0 $

whose positive solution is

$ a = \dfrac{1}{2} ( 2 + \sqrt{12} ) = 1 + \sqrt{3} $

hence,

$ b = a + 1= 2 + \sqrt{3} $

Now,

$ F + G + H = \sqrt{5} (3 + \sqrt{3}) $

And

$ FGH = (1 + \sqrt{3}) (2 + \sqrt{3}) x (5 - x^2) = \dfrac{1}{2} (5 + 3 \sqrt{3} ) A$

Hence, the AM-GM inequality we have

$ \left( \dfrac{ \sqrt{5} (3 + \sqrt{3}) }{3} \right) \ge \sqrt[3]{\dfrac{1}{2} (5 + 3 \sqrt{3} ) A }$

Therefore,

$ A \le \dfrac{2}{5 + 3 \sqrt{3}} \left(\dfrac{ \sqrt{5} (3 + \sqrt{3}) }{3} \right)^3 $

Numerically, this is

$ A \le 8.606629658424 $

i.e.

$ \text{Maximum Rectangle Area } = 8.606629658424 $

This is the same value obtained in the other answer that uses calculus.

The value of $x$ at which this happens is given by

$x = \dfrac{\sqrt{5}}{a - 1} = \sqrt{\dfrac{5}{3}} $

using this, we can express the maximum area as follows

$ \text{ Maximum Rectangle Area } = 2 x (5 - x^2) = 2 \sqrt{ \dfrac{5}{3} } \left(\dfrac{10}{3} \right) = \dfrac{ 20 \sqrt{5}}{3 \sqrt{3}} $

Answer 2

As I could not personally find a solution using only pre-calculus, this answer uses calculus:

The formula for the area of this rectangle will be $A(x)=(2x)(5-x^2)$ where we define x as the positive corner of the rectangle that intersects the parabola.
The formula is $A(x)=(2x)(5-x^2)$ because a rectangle's formula for area is $A=LW$ because the width will be the height of the function and the length will be twice x, and as the function is even, x and -x will be the same height. $0<x<\sqrt5$ because if $|x|>\sqrt5$ then $5-x^2$ will be less than 0, so it will not be above the x axis.

To approach this question, we can use calculus. So we are trying to find the maximum of $(2x)(5-x^2)$ or equivalently $-2x^3+10x$ on the interval $(0,\sqrt5)$. To do this, we must analyze the critical points of the derivative, $A'(x)$ and the endpoints of the interval.
$A'(x)=-6x^2+10$
Then to find critical points, set $A'(x)=0$
$-6x^2+10=0$
$x^2=\frac {10}6$
$ x= \pm \sqrt {\frac {5}3}$
We are looking on the interval $(0,\sqrt5)$ so we will disregard the critical point at $x=-\sqrt {\frac {5}3}$. Our largest area will either be at the critical point or at the endpoints of the interval, so we will test those three cases: $x=0, \sqrt {\frac {5}3}, \sqrt {5}$
$A(0)=(2(0))(5-(0)^2)=0$
$A(\sqrt {\frac {5}3})=(2(\sqrt {\frac {5}3}))(5-(\sqrt {\frac {5}3})^2)=(2\sqrt {\frac {5}3})({\frac {10}3})=\frac {20\sqrt {\frac {5}3}} 3≈8.60662965824$
$A(\sqrt 5)=(2(\sqrt 5))(5-(\sqrt 5)^2)=0$
Of these, $A(\sqrt {\frac {5}3})$ is greatest, so it is the maximum.
So the maximum area is about $8.607$, the base is twice x, or $2\sqrt {\frac {5}3}$, and the height is $5-{\frac {5}3}$ or $\frac {10}3$.