Asked by Sal
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
A) 108
B) 27
C) 12 root 3
D) 54
E) 24 root 3
Answers
Answered by
K
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.
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.
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