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The local theatre has a full house of 1050 nightly and charges $15 per ticket. The manager estimates that the ticket sales would decrease by 50 for each $1 increase in the ticket cost. What ticket price should be charged for maximum income?
Last edited by Identity (2007-08-06 01:07:49)
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Denoting the price increase as x, the amount of tickets sold is given by (1050-50x), or 50(21-x) if you prefer, and the price per ticket is (15+x).
The total revenue is then 50(21-x)(15+x).
Differentiate this and equate to 0 to find the maximum.
Why did the vector cross the road?
It wanted to be normal.
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Thanks mathsyperson
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Be careful because if you differentiate, and set to zero, you might be finding the minimum. But since it is x * -x, that parabola open down, so it probably is a maximum, sorry for confusion.
igloo myrtilles fourmis
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in the event that parabola opened up, the question would have been impossible since the maximum would be undefined
Last edited by luca-deltodesco (2007-07-26 02:46:49)
The Beginning Of All Things To End.
The End Of All Things To Come.
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There is no need to differentiate. Since the object function is quadratic, just complete the square.
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