Question

select all the correct answers
a farmer wants to build two fenced - off sections within his field, one in the shape of a rectangle and the other in the shape of a square. the side of the square must be equal to the width of the rectangle, x feet. the length of the rectangle must be 50 feet longer than its width.
the field the farmer wants to build the two fenced sections in has an area of y square feet. the difference of the area of this field and the area of the fenced, square section needs to be at least 1,000 square feet. in addition, the sum of the fenced areas must be less than the area of the field.
this is the system of inequalities that represents this situation.
$y\geq x^{2}+1,000$
$y > 2x^{2}+50x$
which points represent viable solutions?
$(30,2,000)$
$(25,2,500)$
$(20,2,200)$
$(5,3,000)$
$(10,1,100)$

Explanation:

Step1: Substitute \(x = 30\) into inequalities

For \(y\geq x^{2}+1000\), \(x^{2}+1000=30^{2}+1000 = 900 + 1000=1900\), and \(y = 2000\geq1900\). For \(y>2x^{2}+50x\), \(2x^{2}+50x=2\times30^{2}+50\times30=2\times900 + 1500=1800+1500 = 3300\), and \(2000<3300\). So \((30,2000)\) is not a solution.

Step2: Substitute \(x = 25\) into inequalities

For \(y\geq x^{2}+1000\), \(x^{2}+1000=25^{2}+1000=625 + 1000 = 1625\), and \(y = 2500\geq1625\). For \(y>2x^{2}+50x\), \(2x^{2}+50x=2\times25^{2}+50\times25=2\times625+1250 = 1250+1250=2500\), and \(2500
ot>2500\). So \((25,2500)\) is not a solution.

Step3: Substitute \(x = 20\) into inequalities

For \(y\geq x^{2}+1000\), \(x^{2}+1000=20^{2}+1000=400 + 1000=1400\), and \(y = 2200\geq1400\). For \(y>2x^{2}+50x\), \(2x^{2}+50x=2\times20^{2}+50\times20=2\times400+1000 = 800 + 1000=1800\), and \(2200>1800\). So \((20,2200)\) is a solution.

Step4: Substitute \(x = 5\) into inequalities

For \(y\geq x^{2}+1000\), \(x^{2}+1000=5^{2}+1000=25 + 1000=1025\), and \(y = 3000\geq1025\). For \(y>2x^{2}+50x\), \(2x^{2}+50x=2\times5^{2}+50\times5=2\times25+250 = 50+250=300\), and \(3000>300\). So \((5,3000)\) is a solution.

Step5: Substitute \(x = 10\) into inequalities

For \(y\geq x^{2}+1000\), \(x^{2}+1000=10^{2}+1000=100 + 1000=1100\), and \(y = 1100
ot\geq1100\) (since \(y\geq x^{2}+1000\) is non - strict inequality here, but also check the second inequality. For \(y>2x^{2}+50x\), \(2x^{2}+50x=2\times10^{2}+50\times10=200 + 500=700\), \(y = 1100>700\), but \(y=x^{2}+1000\) at \(x = 10,y = 1100\) and the first inequality is \(y\geq x^{2}+1000\) (equality holds). However, in the context of areas (dimensions \(x>0\)), we can also check another aspect. The side of the square and width of rectangle \(x>0\). But if we consider the non - strict inequality \(y\geq x^{2}+1000\) and \(y>2x^{2}+50x\), when \(x = 10,y=1100\), \(2x^{2}+50x=2\times100 + 500=700\), \(1100>700\) and \(1100=10^{2}+1000\). But usually, for the sum of fenced areas (\(x^{2}+x(x + 50)=2x^{2}+50x\)) and field area \(y\), if \(y=x^{2}+1000\) and \(2x^{2}+50x0\) and non - degenerate shapes. Also, if we consider the physical meaning of the problem (dimensions of the rectangle and square, \(x>0\)). But if we just substitute into the inequalities:
\(y\geq x^{2}+1000\): \(1100=10^{2}+1000\)
\(y>2x^{2}+50x\): \(1100>2\times10^{2}+50\times10=700\). But since \(x\) is a length (width of rectangle and side of square), \(x>0\). However, if we consider the non - strict inequality \(y\geq x^{2}+1000\) (it's allowed). But another way: the sum of the fenced areas is \(A_f=2x^{2}+50x\) and \(y - x^{2}\geq1000\) (i.e., \(y\geq x^{2}+1000\)). When \(x = 10\), \(A_f=2\times100+500 = 700\), \(y=1100\). But if we assume that the farmer is building non - degenerate fenced areas (a rectangle with \(x>0\) and length \(x + 50>x\)), \(x = 10\) gives a valid rectangle (length \(x+50=60\)) and square (side \(x = 10\)).

Answer:

\((20,2200)\), \((5,3000)\)