The intersection of the graphic solutions of all three inequalities listed below form an enclosed region in the coordinate plane. What is the area of this region? Express your answer in terms of $\pi$.

\(\[ \left\{ \begin{array}{l} (x-4)^2+y^2\leq 16 \\ y \geq x-4 \\ y\geq -\frac{1}{3}x \end{array} \right. \]\)

michaelcai
Nov 2, 2017

#1**+2 **

Look at the graph, here : https://www.desmos.com/calculator/ekrnfonsjr

We have two areas to consider

One is a triangle with a base of 4 and a height of 1.....so...its area is 2 units^2

The other area is 3/4 of a semi-circle with a radius of 4

And the area of this is (1/2)(3/4) pi (4)^2 = 6 pi units^2

And the feasible region will be the sim of these = [ 6pi + 2] units^2

CPhill
Nov 2, 2017