Re: isi2010

Posted by neha on Apr 24, 2012; 4:04pm
URL: http://discussion-forum.276.s1.nabble.com/isi2010-tp7496231p7496444.html

solution 1:

I think it should be done as follows:

The profit equation will be
   x^2 -60x+800 =0 which gives x=20 or x=40,,,
Now the revenue equation is
               TR(x)= (100-x)x/100
             
Since there are only two values of x for which the profit constraint is satisfied,,,
       TR(20)=160 and TR(40)=240,, so the revenue is maximum when x is 40 so price will be 6 per unit...