Answer:
The maximum area of the triangular region is 2 square units
Step-by-step explanation:
we know that
The area of the triangular region is equal to
[tex]A=\frac{1}{2}(x)(y)[/tex] -----> equation A
where
x is the x-coordinate of point P
y is the y-coordinate of point P
we have
The equation of the graph of the line is
[tex]x+y=4[/tex]
[tex]y=-x+4[/tex] -----> equation B
substitute equation B in equation A
[tex]A=\frac{1}{2}(x)(-x+4)[/tex]
[tex]A=-\frac{1}{2}x^2+2x[/tex]
This is a vertical parabola open downward (the leading coefficient is negative)
The vertex represent a maximum
The y-coordinate of the vertex represent the maximum area of the triangular region
Find the vertex
Convert the quadratic equation in vertex form
Factor -1/2
[tex]A=-\frac{1}{2}(x^2-4x)[/tex]
Complete the square
[tex]A=-\frac{1}{2}(x^2-4x+4)+2[/tex]
Rewrite as perfect squares
[tex]A=-\frac{1}{2}(x-2)^2+2[/tex]
The vertex is the point (2,2)
so
The maximum area is the y-coordinate of the vertex
The maximum area of the triangular region is 2 square units
Find the coordinates of point P for the maximum area
x=2 (x-coordinate of the vertex)
Find the y-coordinate of point P
substitute in equation B
y=-2+4=2
The coordinates of point P for the maximum area is P(2,2)
