calculus
posted by ej on .
a right triangle is formed in the first quadrant by the x and y axes and a line through the point (1,2). write the length of the hypotenuse as a function of x. find the vertices of the triangle such that its area is a minimum.

let P(x,0) be the xintercept and Q(0,y) be the yintercept
label (1,2) as A(1,2)
the slope AP = slope AQ
2/(x1) = (2y)/1
2x  xy  2 + y = 2
y(1x) = 2x
y = 2x/(1x) or 2x/(x1)
H^2 = x^2 + y^2
= x^2 + ( (2x/(x1) )^2  simplify if needed
(don't know why we are finding the hypotenuse ? )
Area of triangle OPQ
= (1/2)base x height
= (1/2)xy
= (1/2)x(2x/(x1))
= (1/2) (2x^2/(x1) )
dA/dx = (1/2) [ (x1)(4x)  2x^2(1) ]/(x1)^2
= (1/2) [ 4x^2  4x  2x^2 ]/(x1)^2
= 0 for a max of A
2x^2  4x = 0
x(2x4) = 0
x=0  > makes no sense, look at diagram
or
x = 2
then y = 2(2)/(21) = 4
the triangle has (2,) and (0, 4) as its vertices for a maximum area 
sticky keyboard, last line should obviously say
the triangle has (2,0) and (0, 4) as its vertices for a maximum area