the area of the largest isosceles triangle that can be drawn with ne vertex at the origin and with the 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 sq rt (3)
D. 54
E. 24 sq rt. (3)
2007-04-14 17:33:25 · 3 answers · asked by jpatel10989 1 in Science & Mathematics ➔ Mathematics
The area of one of these triangles would be: (2x)=base times (27-x^2)=height, divided by two. Therefore, the area formula that you want to maximize is x(27-x^2)=27x-x^3. The derivative is 27-3x^2; set this to zero and obtain: x^2=9 which means that the largest area will occur when the other two points of the triangle are located at points (3,18) and (-3,18). This means that your base is 6 and your height is 18; therefore your area is 6*9=54.
2007-04-14 17:42:40 · answer #1 · answered by bruinfan 7 · 0⤊ 0⤋
make "generic" point on graph call it (x , sqrt(27-x^2) )
Area of triangle = 1/2 * base * height
A = (1/2) * (2x) (27-x^2) = x(27-x^2) = 27x - x^3
to find maximum we set A' = 0
A' = 27 - 3x^2
set equal to zero
27-3x^2 = 0
x^2 = 9
x = +/- 3
plug into A = 27x - x^3
A = 81 - 27 = 54
:)
2007-04-15 00:44:55 · answer #2 · answered by Anonymous · 0⤊ 0⤋
c
2007-04-15 00:40:33 · answer #3 · answered by Anonymous · 0⤊ 0⤋