Answer:
a) The yield for the circular area is the triple of the original yield so it is 1,050 bushels.
Average yield for the square plot= 899 bushels.
b) The ratio of irrigated area to the total area of the square plot is the same if we put the sprinkler in the center (with half side as the ratio) or in the corner (with ratio of the side of the square).
Step-by-step explanation:
The arm of the sprinkler goes half the length of the square plot, so the area irrigated is the circumscribed circle. See the picture attached (irrigated area).
The yield will triple in the irrigated area and the same as before in the non-irrigated area. The irrigated area is the area of the circle with radius equal to half the side of the square.
[tex]Y_c=3Y_0=3\cdot350=1,050[/tex]
The non-irrigated area is the area of the square less the area of the circle.
We can calculate the yield as:
[tex]Y=\dfrac{(3Y_o)\cdot A_c}{A_s}+\dfrac{Y_0(A_s-A_c)}{A_s}[/tex]
where Y0: original yield (350 bushels), Ac: area of the circle, and As: area of the square.
The ratio between the area of the circle and the area of the square is:
[tex]\dfrac{A_c}{A_s}=\dfrac{\pi r^2}{l^2}\\\\\\r=l/2\\\\\\\dfrac{A_c}{A_s}=\dfrac{\pi (l/2)^2}{l^2}=\dfrac{\pi l^2}{4l^2}=\dfrac{\pi}{4}\approx0.78[/tex]
We can now solve for the yield:
[tex]Y=\dfrac{(3Y_o)\cdot A_c}{A_s}+\dfrac{Y_0(A_s-A_c)}{A_s}\\\\\\Y=3Y_0\dfrac{\pi}{4}+Y_0(1-\dfrac{\pi}{4})=Y_0\left(\dfrac{3\pi}{4}+1-\dfrac{\pi}{4}\right)=Y_0\left(\dfrac{\pi}{2}+1\right)\\\\\\Y\approx2.57Y_0=2.57\cdot350=899.5[/tex]
b) If the arm is located in one corner of the square and the arm has a length of the side of the square, we have an irrigated area that is a quarter of a circle with radius l.
We can calculate the proportion of irrigated area in the square plot, and if it is bigger than the proportion of irrigated area when we have the sprinkler in the center (proportion of 0.78 of irrigated area), then it is better to put the sprinkler in the corner.
The ratio of the irrigated area to the total area is:
[tex]\dfrac{Aq}{A_s}=\dfrac{\pi r^2/4}{l^2}\\\\\\r=l\\\\\dfrac{Aq}{A_s}=\dfrac{\pi l^2/4}{l^2}=\dfrac{\pi}{4}\approx 0.78[/tex]
The ratio of irrigated area to the total area of the square plot is the same if we put the sprinkler in the center (with half side as the ratio) or in the corner (with ratio of the side of the square).
