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Word problems

May 14, 2019

Grade 10s are working on quadratic word problems. We’re looking at maximum areas for fields, and maximum revenue for bushels of potatoes. In any situation we need to first identify variables, make an equation, or two, with the goal to make something quadratic. We are looking here for the maximum area or revenue, so we want to find the vertex (sommet) so we need to remember all the ways to do that, and hopefully choose a way that is quick. It’s often easy to find the zeros, the midpoint, and then the vertex.our revenue problem involved a farmer who started with 600 bushels of potatoes to be sold at 2$/ bushel. If she waits, she gets 100 more bushels a week, but the price drops by 10 cents a bushel. She needs to know when to harvest, to get the highest revenue.

We can make a table and solve, but there are sometimes situations where a table becomes a very long tedious method. From the table we notice it is quadratic data. We notice that revenue is equal to price times number of items sold, and we also can make expressions for price and number based on waiting x number of weeks.using the table we see the maximum revenue occurs at 7 weeks. We also see the y intercept is 1200$. If we expand to make the trinomial we see the constant is 1200. We also noticed something neat. The 2nd differences are always double the “a” value of our quadratic. We can use the trinomial form and the quadratic formula, or go back to the factored form and solve for the zeros, we could also complete the many ways to get to the vertex! We find the ridiculous cases (the “stupid” case) where revenue is equal to zero. In this question if the farmer waits 20 weeks there will be lots of potatoes to sell but the price will have dropped to zero. If we sold 6 weeks ago the price would have been higher, but there would have been no potatoes to sell! we look to the graph. We can plot those ridiculous cases with 0 revenue, and then find the sweet spot in the middle where the revenue is maximized. We calculate the midpoint to be 7 and then can calculate the revenue when x is 7 by using our formula.

It’s a bit of a surprise for some of us to be seeing parabolas appear in problems that do not involve kicking a ball. We are seeing them in area problems, price problems, and patterning too. We follow the same steps as we’ve done before, but now need to keep a context in mind and make good choices of the path through the problem.

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