Question

PLEASE HELP! I REALLY NEED THESE ANSWERS!

A right triangle is formed in the first quadrant by the x- and y- axes and a Straight line through the point (2,3). The other vertices are (0,0) (0,Y) (X,0).

a) Write the length L of the hypotenuse as a function of x
b) Find the vertices of the triangle such that its are is a minimum

Answer 1

Heyyyy

Explanation:

Answer 2

Answer:the triangle has (2,) and (0, 4) as its vertices for a maximum area

Step-by-step explanation:let P(x,0) be the x-intercept and Q(0,y) be the y-intercept

label (1,2) as A(1,2)

the slope AP = slope AQ

-2/(x-1) = (2-y)/1

2x - xy - 2 + y = -2

y(1-x) = -2x

y = -2x/(1-x) or 2x/(x-1)

H^2 = x^2 + y^2

= x^2 + ( (2x/(x-1) )^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/(x-1))

= (1/2) (2x^2/(x-1) )

dA/dx = (1/2) [ (x-1)(4x) - 2x^2(1) ]/(x-1)^2

= (1/2) [ 4x^2 - 4x - 2x^2 ]/(x-1)^2

= 0 for a max of A

2x^2 - 4x = 0

x(2x-4) = 0

x=0 ---- > makes no sense, look at diagram

or

x = 2

then y = 2(2)/(2-1) = 4

the triangle has (2,) and (0, 4) as its vertices for a maximum area