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.
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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