Question

Point p(a,b) is in the first quadrant on the graph of the line x+y=4. A triangular region is shown on the diagram. What is the maximum area of the triangular region?

Answer 1

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

$A=\frac{1}{2}(x)(y)$ -----> 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

$x+y=4$

$y=-x+4$ -----> equation B

substitute equation B in equation A

$A=\frac{1}{2}(x)(-x+4)$

$A=-\frac{1}{2}x^2+2x$

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

$A=-\frac{1}{2}(x^2-4x)$

Complete the square

$A=-\frac{1}{2}(x^2-4x+4)+2$

Rewrite as perfect squares

$A=-\frac{1}{2}(x-2)^2+2$

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)