A right triangle has one vertex on the graph of y = 8 - 2x, x > 0 at (x, y), another at the origin, and the third on the positive y-axis at (0,y)

Question

A. Express the area A of the triangle as a function of x.

B. For what value of x is A the largest?

C. What is the maximum area?

Answer 1

see answer from Yelfim, expert tutor. Note, one side on y-axis, one side is from origin

to line 8-2x, and third side is parallel to x-axis connecting vertex on y-axis and vertex

on line.

Answer 2

two basic right triangles are possible.

the one with max Area is when x=0,

max Area = 16

top of the right triangle is the point (x,y) = (0,8) , right vertex is (4,0), which with (0,0)

create a right triangle with Area = 32/2 = 16

the point (3 1/3, 1 1/3) or (10/3, 4/3) also creates a right triangle with vertices (4,0) and (0,0) with

area = (1/2)[(3 1/3)(1 1/3) + (2/3)(1 1/3)

= (1/2)(1 1/3)(3 1/3+2/3)

=(1/2)(2/3(4) = 8/3 = 2 2/3 < 16

Largest Area is 16 when x= 0

A= f(x) = (1/2)hb where h=height of the triange and b= base

=(1/2)(8)(4) = 32/2 =16

and

(1/2)(1 1/3)(4)= (1/2)(4/3)(4) = (8/3) = 2 2/3

A(x) = f(x)= 16 -4x

f(0) = 16-4(0) = 16

f(3 1/3) = 16-4(10/3) = 48/3 -40/3 = 8/3 = 2 2/3

x= slightly, infinitesmally larger than zero gives the largest area, approximately virtually 16

A(x) = 16-4x is the Area function of x

but that's only if y is required to be positive

if y has no restrictions

the largest Area is when y= - infinity, with Area = infinity

plotting the line and points help a lot

then

the area equation becomes

A(x) = |16-4x|= the abolute value of (16-4x)