# 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,

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^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 = 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

Instead we can observe that a maximum in

# 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

# (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

When

# y = sqrt(529-x^2) #

# \ \ = sqrt(529-529/2) #

# \ \ = sqrt(529/2) #

# \ \ = (23sqrt(2))/2 #

Thus the critical point correspond to

We should really validate that this critical point corresponds to a maximum., Given the earlier alternative answer

Differentiating our earlier result we have:

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

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

And with

# (d^2Phi)/dx^2 = 1058-12*529/2 #

# " " = -2116 #

# " " lt 0 => # maximumHence the maximum area enclosed by

#triangle OPQ# occurs when#x=y# QED

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so that:

The area of such triangle is:

and evaluating the derivative:

and then:

The maximum area is:

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To prove that the maximum area occurs when OP=OQ, we need to use the formula for the area of a triangle formed by connecting three points in a coordinate plane. The area (A) of a triangle formed by points (x1, y1), (x2, y2), and (x3, y3) is given by:

A = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Given that P and Q lie on the y-axis, their x-coordinates are both 0. Let P be (0, p) and Q be (0, q), where p and q are the y-coordinates of P and Q respectively.

Therefore, the coordinates of P and Q are (0, p) and (0, q) respectively.

The distance between P and Q (PQ) is given as 23, so |p - q| = 23.

We want to maximize the area of triangle OPQ. Since O is the origin, the coordinates of O are (0, 0).

Substituting the coordinates of O, P, and Q into the area formula:

A = 1/2 * |0(p - q) + 0(0 - p) + 0(q - 0)|

A = 1/2 * |0 + 0 + 0|

A = 0

Since the area of a triangle cannot be negative, the maximum area occurs when the expression inside the absolute value is maximized.

To maximize |p - q|, p and q should be equal.

Therefore, the maximum area occurs when OP = OQ.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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