maths help

Question

Find the area of the region described by y≥ 2|x-3|+4 and x-2y ≥ -20.

Answer 1

Shivam,

 

I don't have time to write it all out right now, but here's how you need to approach the problem.

 

You have to solve absolute value equations in two parts.

 

When (x-3) ≥ 0 you get one line.  That's when x ≥ 3.

  For x ≥ 3, y = 2(x-3)+4

  y ≥ 2(x-3)+4

 

When (x-3) ≤ 0 you get a different line. That's when x ≤ 3.

  For x ≤ 3, y = -2(x-3)+4            (x-3) will always be negative but |x-3| will be positive!

This means that |x-3|=-(x-3)

   y ≥ -2(x-3)+4

 

The line x-2y ≥ -20 should intersect both of those lines to form a triangle.

Solve this inequality:

-2y ≥ -x-20

2y ≤ x + 20           You have to reverse the inequality when you multiply or divide by a negative (-1 in this case).

y ≤ 1/2 x + 10

 

Draw a graph of these three inequalities to see the triangle that they define.

You'll have to find all of the line intersections graphically or as systems of equations.  Each line is defined by an equation with "y=...", so:

 

For the intersection of y = 2(x-3)+4 and y = 1/2 x + 10, just do this:

2(x-3)+4 = 1/2 x + 10

Solve for x, plug that back into one of these two equations to calculate y.

The point (x, y) is that intersection.

 

You'll probably have to do a little more work to find the length of a base and the height of the triangle so you can calculate its area.