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Pre-Calc Homework HELP!?
A right triangle has one vertex on the graph of y=9-x^2, x>0, at (x,y), another at the origin, and a third on the positive x-axis at (x,0). Express the area A of a triangle as a function of x. Find maximum area.
Any help would be awesome!
1 Answer
- ?Lv 69 years agoFavorite Answer
sketch the graph and yout will find a parabola with intersect points (0,9) and (3,0)
Take any point (x,y) on the graph and draw the triangle by droping a perpendicular to the x-axis.
The area of the triangle is base * height * 1/2,
base is the x-coordinate,
height is the y- coordinate
therefore A = x *y * (1/2) = x*(9-x^2) * (1/2) = 9x/2 - (x^3)/2
to find the maximimum, calculate dA/dx and then equate to zero to find the x-value of the maximum point, then put maximum value into the equation for A. (NB remember we are only considering positive values of x)