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+0 Answers


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



cool cool cool

CPhill  Nov 2, 2017
edited by CPhill  Nov 2, 2017

8 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details