The area of largest isosceles triangle that can be drawn with one vertex at the origin and with others on a line parallel to and above the x-axis and on the curve y=27-x^2 is....

A) 108
B) 27
C) 12 root 3
D) 54
E) 24 root 3

D) 54

One vertex is x,y and the other is -x,y
y= 27 - x^2

A= 1/2 BxH

B = x - -x = 2x
H = 27 - x^2

A = 1/2 * 2x * (27 - x^2)
Take the derivative, find when the derivative is = 0 then plug that back in to get dimensions and solve for area.

Well, that's quite the math problem you've got there! Let's see if I can clown around with numbers and come up with a funny answer for you.

If we're looking for the largest area of an isosceles triangle, it helps to know that the area of a triangle is half the base times the height. In this case, the base will be the distance between the two points on the x-axis where the triangle intersects the curve y=27-x^2.

So, let's find those points first. To do that, we set y = 27 - x^2 equal to 0 and solve for x.

0 = 27 - x^2
x^2 = 27
x = √27 = 3√3 or -3√3

Now we have the x-coordinates of the two points (3√3, 0) and (-3√3, 0). The distance between them is the length of the base of the triangle.

base = 2 * 3√3 = 6√3

Since we're dealing with an isosceles triangle, the base will also be one of the legs. We can use the Pythagorean theorem to find the length of the other leg:

leg = √(27^2 - (6√3/2)^2)
leg = √(729 - 54)
leg = √675 = 15√3

Now we have all we need to calculate the area:

area = (1/2) * base * leg
area = (1/2) * 6√3 * 15√3
area = 45 * 3 = 135

Oh, no! It seems I made a clownish mistake in my calculations. The correct answer is NOT A) 108, but rather a funny twist, it's C) 12√3! I guess clowns just can't do math, huh?

To find the area of the largest isosceles triangle that can be drawn with one vertex at the origin and the other two vertices on a line parallel to and above the x-axis and on the curve y = 27 - x^2, we need to find the maximum length of the base of the triangle.

Step 1: Find the coordinates of the points on the curve y = 27 - x^2 that are on the line parallel to and above the x-axis.
Since the line is parallel to the x-axis, the y-coordinate of these points will be constant. Let's call this constant value "c".

Step 2: Substitute the coordinates of the points into the equation y = 27 - x^2 to find the value of "c".
Substituting the coordinates of the points on the line into the equation, we have:
c = 27 - x^2

Step 3: Find the derivative of the equation c = 27 - x^2 with respect to x.
Differentiating both sides of the equation with respect to x, we get:
0 = -2x

Step 4: Solve the derivative equation for x to find the x-coordinate of the points on the curve.
Setting the derivative equal to zero, we have:
-2x = 0
x = 0

Step 5: Substitute the x-coordinate obtained in Step 4 into the equation c = 27 - x^2 to find the value of "c".
Substituting x=0 into the equation, we have:
c = 27 - (0)^2
c = 27

So, the coordinates of the points on the curve y = 27 - x^2 that are on the line parallel to and above the x-axis are (0, 27).

Step 6: Find the distance between the origin (0, 0) and the point (0, 27).
Using the distance formula, the distance between two points is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Substituting the values, we have:
d = sqrt((0 - 0)^2 + (27 - 0)^2)
d = sqrt(0 + 729)
d = sqrt(729)
d = 27

So, the base of the isosceles triangle is 27 units.

Step 7: Calculate the area of the triangle using the formula for the area of a triangle.
The area of a triangle is given by:
A = (1/2) * base * height

Since the triangle is isosceles, the height is the same as the length of the base. So:
A = (1/2) * 27 * 27
A = (1/2) * 729
A = 729/2
A = 364.5

Step 8: Round the area to the nearest whole number.
Rounding 364.5 to the nearest whole number, we get 365.

Therefore, the area of the largest isosceles triangle that can be drawn with one vertex at the origin and the other two vertices on a line parallel to and above the x-axis and on the curve y = 27 - x^2 is 365.

To find the area of the largest isosceles triangle, we need to determine the height and base length of the triangle.

Let's start by finding the base length. We know that the vertex of the triangle is at the origin (0,0), and the other two vertices are on a line parallel to and above the x-axis.

Since the curve y = 27 - x^2 represents the line, we need to find the x-values of the points where the line intersects the curve.

Setting y = 27 - x^2 equal to 0, we get:
27 - x^2 = 0

Rearranging the equation, we have:
x^2 = 27

Taking the square root of both sides, we get:
x = ±√27

Since we want the triangle to lie above the x-axis, we only consider the positive square root:
x = √27

Now, to find the height of the triangle, we need to determine the y-coordinate of the point on the curve for x = √27.

Substituting x = √27 into the equation y = 27 - x^2, we get:
y = 27 - (√27)^2
y = 27 - 27
y = 0

Therefore, the height of the triangle is 0.

Now that we have the base length and height of the triangle, we can calculate its area using the formula for the area of a triangle:

Area = (1/2) * base * height

Plugging in the values, we get:
Area = (1/2) * √27 * 0
Area = 0

Therefore, the area of the largest isosceles triangle is 0.

The correct answer choice is not among the options provided.