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#?

Answer 1

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

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Answer 2
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#
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Answer 3

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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Answer from HIX Tutor

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.

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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