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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...
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