How to Applying Quadratics to the Real World

Quadratic functions are wonderful models for many situations that occur in the real world. You can see them at work in financial and physical applications, just to name a couple. This section provides a few applications for you to consider.

Selling candles

A candle-making company has figured out that its profit is based on the number of candles it produces and sells. The function P (x) = –0.05x2 + 8x – 140 applies to the company’s situation, where x represents the number of candles, and P represents the profit. You can use the function to find out how many candles the company has to produce to garner the greatest possible profit.

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You find the two x-intercepts by letting y = 0 and solving for x by factoring:

0 = -0. 05x2 + 8x – 140

0 = -0. 05(x2 – 160x + 2800)

0 = -0. 05(x – 20) (x – 140)

x – 20 = 0 or x – 140 = 0

x = 20 or x = 140

The intercept (20, 0) represents where the function (the profit) changes from negative values to positive values. You know this because the graph of the profit function is a parabola that opens downward (because a is negative), so the beginning and ending of the curve appear below the X-axis. The intercept

(140, 0) represents where the profit changes from positive values to negative values. So, the maximum value, the vertex, lies somewhere between and above the two intercepts.

You now use the formula for the x-coordinate of the vertex, x=\frac{-b}{2a}, to find x=\frac{-8}{2(-0.05)}=\frac{-8}{-0.1}=80. The number 80 lies between 20 and 140; in fact, it rests halfway between them. The nice, even number is due to the symmetry of the graph of the parabola and the symmetric nature of these functions. Now you can find the P value (the y-coordinate of the vertex): P (80) = –0.05(80)2 + 8(80) – 140 = –320 + 640 – 140 = 180.

Your findings say that if the company produces and sells 80 candles, the maximum profit will be $180. That seems like an awful lot of work for $180, but maybe the company runs a small business. Work such as this shows you how important it is to have models for profit, revenue, and cost in business so you can make projections and adjust your plans.

Shooting basketballs

A local youth group recently raised money for charity by having a Throw-AThon. Participants prompted sponsors to donate money based on a promise to shoot baskets over a 12-hour period. This was a very successful project, both for charity and for algebra, because you can find some interesting bits of information about shooting the basketballs and the number of misses that occurred.

Participants shot baskets for 12 hours, attempting about 200 baskets each hour. The quadratic equation M(t)=\frac{17}{6}t^2-\frac{77}{3}t+100 models the number of baskets they missed each hour, where t is the time in hours (numbered from 0 through 12) and M is the number of misses.

From the graph, you see that the initial value, the y-intercept, is 100. At the beginning, participants were missing about 100 baskets per hour. The good news is that they got better with practice. M (2) = 60, which means that at hour two into the project, the participants were missing only 60 baskets per hour. The number of misses goes down and then goes back up again. How do you interpret this? Even though the participants got better with practice, they let the fatigue factor take over.

Also Read : Solve revenue and distance applications of quadratic equations

What’s the fewest number of misses per hour? When did the participants shoot their best? To answer these questions, find the vertex of the parabola by using the formula for the x-coordinate.

The best shooting happened about 4.5 hours into the project. How many misses occurred then? The number you get represents what’s happening the entire hour — although that’s fudging a bit:

The fraction is rounded to two decimal places. The best shooting is about 42 misses that hour.

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