Question

A study of how much corn was produced in a field depending on the planting density compared the yield from fields planted with 40000 plants per hectare to the yield from fields planted with 60000 plants per hectare.
1. What is the ratio of the distance between the corn plants in the more-densely-planted field to the distance between the corn plants in the less-densely-planted field?

Answer 1

Answer:

$\frac{d_2}{d_1} =\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}= 1.225$

Or in the other way:

$\frac{d_1}{d_2} =\frac{\sqrt{4}}{\sqrt{6}}=\sqrt{\frac{4}{6}} = \frac{\sqrt{6}}{3}= 0.816$

Explanation:

For this case we need to remember that hectare is a measure of area. And we have the following ratios:

$r_1 =40000 \frac{plants}{acre}$

We can convert this into $plants/m^2$

$r_1 = 40000 \frac{plants}{acre} *\frac{1 acre}{10000 m^2}=4 \frac{plants}{m^2}$

$r_2=60000 \frac{plants}{acre}$

We can convert this into $plants/m^2$

$r_2 = 60000 \frac{plants}{acre} *\frac{1 acre}{10000 m^2}=6 \frac{plants}{m^2}$

By dimensional analysis we assume that $A = L^2$ we can find the ratio in terms of distance like this:

$d_1 =\sqrt{4}= 2 \frac{\sqrt{plants}}{m}$

$d_2= \sqrt{6}= 2.449 \frac{\sqrt{plants}}{m}$

And if we find the ratios in terms of distance we got:

$\frac{d_2}{d_1} =\frac{\sqrt{6}}{\sqrt{4}}=\sqrt{\frac{6}{4}} = \frac{\sqrt{6}}{2}= 1.225$

Or in the other way:

$\frac{d_1}{d_2} =\frac{\sqrt{4}}{\sqrt{6}}=\sqrt{\frac{4}{6}} = \frac{\sqrt{6}}{3}= 0.816$