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`?

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:

`{ (y, y"-Coordinate of the point P"), (x, x"-Coordinate of the point Q"),(A, "Area enclosed by the triangle OPQ") :}`

With all variables being positive. We are given that `PQ=23`, and by Pythagoras

`\ \ \ PQ^2 = AP^2+OQ^2`
`:. 23^2 = y^2+x^2`

`:. x^2+y^2 = 529 => y = sqrt(529-x^2) .... (star)`

And the Area ,`A`, of `triangle OPQ` is given by:

`A = 1/2(x)(y)`
`\ \ = 1/2xsqrt(529-x^2)`

Now this is where a little common sense can make life easier. We could use this function and find `(dA)/dx`, and find a critical point where `(dA)/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 `alphaA^2`, We have from earlier:

`A = 1/2xsqrt(529-x^2) => 2A = xsqrt(529-x^2)`

so instead let us look at the function:

`Phi = (2A)^2`
`\ \ \ = (xsqrt(529-x^2))^2`
`\ \ \ = x^2(529-x^2)`
`\ \ \ = 529x^2-x^4`

Then differentiating wrt `x` we get:

`(dPhi)/dx = 1058x-4x^3`

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

`(dPhi)/dx = 0 => 1058x-4x^3 = 0`
`:. x(1058-4x^2) = 0`
`:. x = 0, 4x^2=1058`
`:. x = 0, x^2=529/2`
`:. x = 0, (23sqrt(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=(23sqrt(2))/2`, we have using `(star)`:

`y = sqrt(529-x^2)`
`\ \ = sqrt(529-529/2)`
`\ \ = sqrt(529/2)`
`\ \ = (23sqrt(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:

`\ \ \ \ \ \ (dPhi)/dx = 1058x-4x^3`

`:. (d^2Phi)/dx^2 = 1058-12x^2`

And with `x=(23sqrt(2))/2`, we have:

`(d^2Phi)/dx^2 = 1058-12*529/2`
`" " = -2116`
`" " lt 0 =>`maximum

Hence the maximum area enclosed by `triangle OPQ` occurs when `x=y` QED

The triangle formed by the points `(0,0)`, `(x.0)`, `(0,y)` 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 = (xy)/2 = (xsqrt(23^2-x^2))/2`

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

`(dS)/dx = 0`

and evaluating the derivative:

`1/2(sqrt(23^2-x^2) -x^2/sqrt(23^2-x^2)) = 0`

`(sqrt(23^2-x^2) -x^2/sqrt(23^2-x^2)) = 0`

`(23^2-x^2 -x^2)/sqrt(23^2-x^2) = 0`

`23^2-2x^2 = 0`

`2x^2= 23^2`

`x = 23/sqrt(2)`

and then:

`y= = sqrt(23^2-x^2) = sqrt(23^2-23^2/2) = sqrt(23^2/2) = 23/sqrt2`

The maximum area is:

`S=(xy)/2= 23^2/4 =132.25`