Maximizing Revenue problem?
When a theater owner charges $5 for admission, there is an average attendance of 180 people. For every $0.10 increase in admission, there is a loss of 1 customer from the average number. What admission should be charged to maximize revenue?
I can't set it up for the life of me. I know part is (180-x) but I can't get the rest.... Anyone know what to do?
Comments
$11.50 with 115 attendees. Maximum revenue is $1322.50.
($5.00 + 0.10n)(180-n) = Maxrev
Run a collumn in Excel 0 through 180. In a second adjacent collumn, insert the above equation operating on the 1st collumn. At 65, the revenue is maxed out at $1322.50. The difference of 180 - n = 115, the number of attendees needed.
we've 2 equations: p=1600-x r=px through fact we are maximizing sales, we could desire to have a function that appears like r= some thing. to that end, what we are able to do, is replace the 1st equation into the 2nd: r=px r= (1600-x)(x) r= 1600x-x^2 with the intention to maximise earnings, we could desire to discover the abolute optimum, or the x cost which produces the in basic terms right y cost. to try this, a) Graph it and hint the max b) Or, take the spinoff and set it equivalent to 0. d/dx(1600x-x^2)=0 1600-2x=0 -2x=-1600 x=-1600/-2 x= 800 to that end, the sales is maximized while the cost is 800 in line with unit.