Question

A farmer has collected data from one of his crops. Suppose the farmer found that the number of seeds, () (in weeks) was
()= t^2/4 seeds per mature plant.
Further, the farmer discovered that only
()=40.56−^2/4+1801.75/^2+188.8/
plants matured to age .
Determine the maximum rate of change of the crop's seed production

Answer 1

The rate of change of the crop's seed production is given by differential of

the function of the number of seeds produced.

• The maximum rate of change of the seed production is approximately 117 seeds per week.

Reasons:

$\displaystyle \mathrm{The \ number \ of \ seeds \ per \ plant, \ S(t)}= \mathbf{\frac{t^2}{4}}$

$\displaystyle \mathrm{Plants \ that \ mature \ to \ age \ t, \ P(t)}= \mathbf{40.56- \frac{t^2}{4} + \frac{1,801.75}{t^2} +\frac{188.8}{t}}$

The number of seeds produced by all the plants, n = S(t) × P(t)

Which gives;

$\displaystyle \mathrm{\ P(t) \times S(t)}= \left(40.56- \frac{t^2}{4} + \frac{1,801.75}{t^2} +\frac{188.8}{t}\right) \times \frac{t^2}{4}$

$\displaystyle \mathrm{ \left(40.56- \frac{t^2}{4} + \frac{1801.75}{t^2} +\frac{188.8}{t}\right) \cdot\frac{t^2}{4}} = \mathbf{-0.0625\cdot t^4+10.14\cdot t^2 + 47.2 \cdot t + 450.4375}$

The maximum rate of change is the highest value of the rate of change function.

To find the maximum rate of change of the crops production, the above

function for the number of seeds is differentiated twice. The first

differentiation is to find the rate of change function, and the second

differentiation is to determine the value of t at the maximum rate of change

as follows;

The rate of change of the crop's seed production is given by following function;

$\displaystyle \frac{d}{dt} \left( -0.0625\cdot t^4+10.14\cdot t^2 + 47.2 \cdot t + 450.4375 \right) = -0.25\cdot t^3+ 20.28 \cdot t - 47.2$

Rate of change function, n'(t) = -0.25·t³ + 20.28·t + 47.2

At the maximum value of the above rate of change function, we have

$\displaystyle \frac{d}{dt} \left( -0.25\cdot t^3+ 20.28 \cdot t - 47.2 \right) = \mathbf{-0.75 \cdot t^2 + 20.28} = 0$

Which gives

$\displaystyle t^2 = \frac{20.28}{0.75} = \mathbf{27.04}$

t = √27.04 = 5.2

At the maximum rate of change, the number of weeks, t = 5.2

Plugging in the value of  t = 5.2 into the rate of change function gives;

n'(5.2) = -0.25·5.2³ + 20.28·5.2 + 47.2 = 117.504

• The maximum rate of change of the crop's seed production is approximately 117 seeds per week.

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