# The point P lies on the y-axis and the point Q lies on the y-axis. A triangle is formed by connecting the origin O to P and Q, If PQ=23 then prove that the maximum area occurs when when OP=OQ?

##### 2 Answers
Apr 25, 2017

Let us start with a picture describing the problem:

Our aim is find the area, $A$, as a function of $x$ alone (or we could equally choose $y$) and maximise the area wrt that variable.

Let us set up the following variables:

$\left\{\begin{matrix}y & y \text{-Coordinate of the point P" \\ x & x"-Coordinate of the point Q" \\ A & "Area enclosed by the triangle OPQ}\end{matrix}\right.$

With all variables being positive. We are given that $P Q = 23$, and by Pythagoras

$\setminus \setminus \setminus P {Q}^{2} = A {P}^{2} + O {Q}^{2}$
$\therefore {23}^{2} = {y}^{2} + {x}^{2}$

$\therefore {x}^{2} + {y}^{2} = 529 \implies y = \sqrt{529 - {x}^{2}} \ldots . \left(\star\right)$

And the Area ,$A$, of $\triangle O P Q$ is given by:

$A = \frac{1}{2} \left(x\right) \left(y\right)$
$\setminus \setminus = \frac{1}{2} x \sqrt{529 - {x}^{2}}$

Now this is where a little common sense can make life easier. We could use this function and find $\frac{\mathrm{dA}}{\mathrm{dx}}$, and find a critical point where $\frac{\mathrm{dA}}{\mathrm{dx}} = 0$, and then validate this corresponds to a maximum, but due to the square root we will end up with a messy derivative and some cumbersome algebra.

Instead we can observe that a maximum in $A$ will also cause a maximum in $\alpha {A}^{2}$, We have from earlier:

$A = \frac{1}{2} x \sqrt{529 - {x}^{2}} \implies 2 A = x \sqrt{529 - {x}^{2}}$

so instead let us look at the function:

$\Phi = {\left(2 A\right)}^{2}$
$\setminus \setminus \setminus = {\left(x \sqrt{529 - {x}^{2}}\right)}^{2}$
$\setminus \setminus \setminus = {x}^{2} \left(529 - {x}^{2}\right)$
$\setminus \setminus \setminus = 529 {x}^{2} - {x}^{4}$

Then differentiating wrt $x$ we get:

$\frac{\mathrm{dP} h i}{\mathrm{dx}} = 1058 x - 4 {x}^{3}$

At a min/max this derivative will be zero; thus:

$\frac{\mathrm{dP} h i}{\mathrm{dx}} = 0 \implies 1058 x - 4 {x}^{3} = 0$
$\therefore x \left(1058 - 4 {x}^{2}\right) = 0$
$\therefore x = 0 , 4 {x}^{2} = 1058$
$\therefore x = 0 , {x}^{2} = \frac{529}{2}$
$\therefore x = 0 , \frac{23 \sqrt{2}}{2}$

We can clearly eliminate $x = 0$ as this would correspond to a $0$ width triangle, thus we have one critical point of interest.

When $x = \frac{23 \sqrt{2}}{2}$, we have using $\left(\star\right)$:

$y = \sqrt{529 - {x}^{2}}$
$\setminus \setminus = \sqrt{529 - \frac{529}{2}}$
$\setminus \setminus = \sqrt{\frac{529}{2}}$
$\setminus \setminus = \frac{23 \sqrt{2}}{2}$

Thus the critical point correspond to $x = y$, as suggested by the question.

We should really validate that this critical point corresponds to a maximum., Given the earlier alternative answer $x = 0$ that we discovered (which obviously corresponds to $A = 0$, a minimal solution), then intuition would suggest the volume varies from this minimum to some maximum. We can confirm this using the second derivative test:

Differentiating our earlier result we have:

$\setminus \setminus \setminus \setminus \setminus \setminus \frac{\mathrm{dP} h i}{\mathrm{dx}} = 1058 x - 4 {x}^{3}$

$\therefore \frac{{d}^{2} \Phi}{\mathrm{dx}} ^ 2 = 1058 - 12 {x}^{2}$

And with $x = \frac{23 \sqrt{2}}{2}$, we have:

$\frac{{d}^{2} \Phi}{\mathrm{dx}} ^ 2 = 1058 - 12 \cdot \frac{529}{2}$
$\text{ } = - 2116$
$\text{ } < 0 \implies$maximum

Hence the maximum area enclosed by $\triangle O P Q$ occurs when $x = y$ QED

Apr 25, 2017

The triangle formed by the points $\left(0 , 0\right)$, $\left(x .0\right)$, $\left(0 , y\right)$ is a square triangle because two sides lay on the $x$ and $y$ axes, which are perpendicular.

Because the hypotenuse of this triangle has length $l = 23$ by hypothesis, we have:

${x}^{2} + {y}^{2} = {23}^{2}$

so that:

$y = \sqrt{{23}^{2} - {x}^{2}}$

The area of such triangle is:

$S = \frac{x y}{2} = \frac{x \sqrt{{23}^{2} - {x}^{2}}}{2}$

The function $S \left(x\right)$ is continuous in the compact interval $x \in \left[0 , 23\right]$ so it admits an absolute maximum. This point is in the interior of the interval as $S \left(0\right) = S \left(23\right) = 0$, while for all other points $S \left(x\right) > 0$, so in such point we have:

$\frac{\mathrm{dS}}{\mathrm{dx}} = 0$

and evaluating the derivative:

$\frac{1}{2} \left(\sqrt{{23}^{2} - {x}^{2}} - {x}^{2} / \sqrt{{23}^{2} - {x}^{2}}\right) = 0$

$\left(\sqrt{{23}^{2} - {x}^{2}} - {x}^{2} / \sqrt{{23}^{2} - {x}^{2}}\right) = 0$

$\frac{{23}^{2} - {x}^{2} - {x}^{2}}{\sqrt{{23}^{2} - {x}^{2}}} = 0$

${23}^{2} - 2 {x}^{2} = 0$

$2 {x}^{2} = {23}^{2}$

$x = \frac{23}{\sqrt{2}}$

and then:

$y = = \sqrt{{23}^{2} - {x}^{2}} = \sqrt{{23}^{2} - {23}^{2} / 2} = \sqrt{{23}^{2} / 2} = \frac{23}{\sqrt{2}}$

The maximum area is:

$S = \frac{x y}{2} = {23}^{2} / 4 = 132.25$