Question 1165122
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Let x be the largest number of medium sizes of which the outlet will have space.


Then from the condition, you have this inequality


    x + {{{(1/2)x}}} + {{{(1/3)x}}} <= 200.


Multiply both sides by 6.  You will get


    6x + 3x + 2x <= 1200,

    11x          <= 1200

      x          <= {{{1200/11}}} = 109.09...


<U>ANSWER</U>.   The largest number of medium sizes of which the outlet will have space is  109, under given conditions.
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Solved.