\section*{Task 1 - 011241}

In 27,000 fertilization trials with phosphorus fertilizers, the following average potato yields were observed:

\[
\begin{array}{cc}
\text { Fertilizer application in terms of P2O5 (dt/ha) } & \text { Yield (dt/ha) } \\
0.0 & 237 \\
0.3 & 251 \\
0.9 & 269
\end{array}
\]

The relationship between the fertilizer application \(x\) (in dt/ha) and the yield \(y\) (in dt/ha) can be approximated by the following relation:

\[
y=a-b \cdot 10^{-k x}
\]

where \(a\), \(b\), and \(k\) are constants.

a) Calculate these constants using the values given above!

b) Calculate the yield for a fertilizer application of 0.6 dt/ha and 1.2 dt/ha!

c) Determine the percentage deviation of the calculated values from the experimentally determined values of \(261 \mathrm{dt} / \mathrm{ha}\) and \(275 \mathrm{dt} / \mathrm{ha}\)!
}

a) It holds that:

\[
\begin{array}{r}
237=a-b \\
251=a-b \cdot 10^{-0.3 k} \\
269=a-b \cdot 10^{-0.9 k}
\end{array}
\]

Equation (1) can be easily rearranged for \(a\) and substituted into the other two equations. This yields:

\[
14=b-b \cdot 10^{-0.3 k} \quad ; \quad 32=b-b \cdot 10^{-0.9 k}
\]

This can be rearranged to:

\[
10^{-0.3 k}=\frac{b-14}{b} \quad ; \quad 10^{-0.9 k}=\frac{b-32}{b}=\left(10^{-0.3 k}\right)^{3}
\]

Substituting both equations into each other, we get:

\[
\begin{array}{r}
\left(\frac{b-14}{b}\right)^{3}=\frac{b-32}{b} \\
\frac{b^{3}-42 b^{2}+588 b-2744}{b^{3}}=\frac{b-32}{b} \\
b^{4}-42 b^{3}+588 b^{2}-2744 b=b^{4}-32 b^{3} \\
0=10 b^{3}-588 b^{2}+2744 b \\
0=10 b \cdot\left(b^{2}-\frac{588}{10} b+\frac{2744}{10}\right)
\end{array}
\]

By inspecting the original equations, it is easy to see that the solution \(b=0\) is invalid. Therefore, it follows that:

\[
\begin{gathered}
b_{1,2}=\frac{294}{10} \pm \sqrt{\frac{86436}{100}-\frac{2744}{10}} \\
b_{1}=53.69 \quad ; \quad a_{1}=290.69 \quad ; \quad k_{1}=0.44 \quad \text { and } \quad b_{2}=5.11 \quad ; a_{2}=242.11
\end{gathered}
\]

For the second set of values, no \(k\) can be calculated. Therefore, this solution is invalid.

The constants thus have the following values: \(b=53.69, a=290.69\) and \(k=0.44\).

b) \(x=0.6 \mathrm{dt} / \mathrm{ha} \rightarrow y=261.46 \mathrm{dt} / \mathrm{ha} ; x=1.2 \mathrm{dt} / \mathrm{ha} \rightarrow y=274.77 \mathrm{dt} / \mathrm{ha}\)

c) \(x=261.46 / 261=1.0018 \rightarrow\) For the first trial, there is a percentage deviation of \(0.18 \%\). \(x=274.77 / 275=0.9992 \rightarrow\) For the second trial, there is a percentage deviation of \(0.08 \%\).