quadratic applications...?

Question

When there are 23 apple trees per acre, the average yield has been found to be 400 apples per tree. For each additional tree planted per acre, the yield per tree decreases by 16 apples per tree. How many additional trees per acre should be planted to maximize the yield? (Round your answer to the nearest whole number.)

Answer 1

There are many ways to approach this question the easiest being if you have a graphing calculator just plug in a few points ie. (23,9200) (24, 9216) into stats and then use the stats operation quadreg and plug that formula into your y values and look at the table for the biggest y or you can look at the graph and use the maxium feature.

if you don't have a graph calc then you need to grab a few points ... I'll use (23,400) (24, 384) (25, 368) (26,352) and (27, 336)... now multiply the x and y for each and then put that in the y value for each x.... ie. (23,9200) (24, 9216) (25, 9200) (26, 9152) (27, 9072)

make make a small t-chart (with 2 extra columns the first d and then d') with your first two as your new x then y points.... with the first extra column(d) take the difference of each y(for the first entry you will not be able to have an answer) ie. 16; -16; -48; -80; now in the second extra column(d') take the difference of the d column and multiply it by -1 ... -32, -32, -32

now you ready to start making your quadratic... take d'(32) and take it to the -2 power then place that number in front of x^2 (I will refer to this number as a)

ok now take this formula ax^2 +bx +K(this is the constant but in this problem you have none) now plug in a value you have for x from your second points and solve for b now this is your quadratic, you should test it on a point or two, but use this to create a huge t-chart of different points and look for the when the value gets smaller, and the point before it gets smaller is your point

we actually ran into the answer while doing the problem which is common in math books but the point (24, 9216) or when you have 24 trees is the maximum yield

Answer 2

Yield per tree with 23 trees Y(23) = 400
Yield per tree with 23+n trees Y(23+n) = 400 - 16*n
Total yield is (yield per tree) times (no of trees) T(n) = (400 - 16*n)*(23 + n)
T(n) = 9200 + 400*n - 368*n - 16*n^2
T(n) = 9200 + 32*n - 16*n^2

Differentiate with resp to n:
T'(n) = 32 - 32*n
For T'(n) = 0, n = 1

So max yield occurs and 23+n = 23+1 = 24 trees