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it's the k of y=a(x-h)^2+k
okay see that basically we have to find where the equation of the profit reaches a maxima. as you can see below, at the maxima or minima the 'slope' of th egraph gets zero, as from that point, it either goes up both ways (for minima) or it goes down both ways (for maxima). |dw:1332405696551:dw| so diffferentiate the equation of the profit to get the slope equation and then equate it with zero. you get the x valur for the maxima of the graph. that 'x' is the optimum no. of pretzels!
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