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Originally Posted by VEN
here's my last question, I hope:
A pizza company's research shows that a 25 cent increase in the price of a pizza results in 50 fewer pizzas being sold. The usual price of $15 for a pizza results in sales of 1000 pizzas. The questions are a) Write the algerbraic expression that models the revenue for this situation. and b) what is the optimum Max value?
so far I wrote down on my sheet y=(15+.25x)(1000-50x)  This models the revenue so I guess I answered the first question
I know I can solve the second question by using a table and writing down the points of the parabola until I reach the max value, but how do I do it using the algebraic method? |
The maximum value of a parabola is found right between the two zeros, so the easy way to derive a formula for the x-value of the maximum, if you know the quadratic formula, is to find the average of the two zeros:
x = ([-b + sqrt(b^2 - 4ac)/(2a)] + [-b - sqrt(b^2 - 4ac)/(2a)]) / 2
This reduces to x = -b/(2a).
So to answer the pizza question you need to multiply out your equation into the form y = ax^2 + bx + c, and apply that formula. That should give you the number of price increases.